Question 14 Marks
Find the point on the curve y = x2 - 2x + 3, where the tangent is parallel to x-axis.
AnswerThe slope of the x-axis is 0
Now, let (x1, y1) be the required point.
Since, the point lies on the curve.
Hence, $\text{y}_1=\text{x}_1^2-2\text{x}_1+3\ ...(1)$
Now, $\text{y}=\text{x}^2-2\text{x}+3$
$\frac{\text{dy}}{\text{dx}}=2\text{x}-2$
Slope of the tangent at (x, y) = $\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(\text{x}_1,\text{y}_1)}=2\text{x}_1-2$
Given:
Slope of the tangent at (x1, y1) = slope of the x - axis
= 2x1 - 2 = 0
⇒ x1 = 1
and
y1 = 1 - 2 + 3 = 2 [From (1)]
$\therefore$ Required point = (x1, y1) = (1, 2)
View full question & answer→Question 24 Marks
Find the points on the curve x2 + y2 = 13, the tangent at each one of which is parallel to the line 2x + 3y = 7.
AnswerThe given equation of curve and the lines is
$\text{x}^2+\text{y}^2=13\ ...(1)$
and $2\text{x}+3\text{y}=7\ ...(2)$
Slope $=\text{m}_1$ for (1)
$\text{m}_1=\frac{\text{dy}}{\text{dx}}=\frac{-\text{x}}{\text{y}}\ ...(3)$
Slope $=\text{m}_1$ for (2)
$\text{m}_2=\frac{\text{dy}}{\text{dx}}=\frac{-2}{3}\ ...(4)$
According to the question
$\text{m}_1=\text{m}_2$
$\Rightarrow\frac{-\text{x}}{\text{y}}=\frac{-2}{3}$
$\Rightarrow\text{x}=\frac{2}{3}\text{y}$
From (1)
$\frac{4}{9}\text{y}^2+\text{y}^2=13$
$\Rightarrow\frac{13\text{y}^2}{9}=13$
$\Rightarrow\text{y}=\pm3$
$\therefore\text{x}=\pm2$
Thus, the point are (2, 3) and (-2, -3).
View full question & answer→Question 34 Marks
If the tangent to the curve y = x3 + ax + b at (1, − 6) is parallel to the line x − y + 5 = 0, find a and b.
AnswerGiven:
x - y + 5 = 0
⇒ y = x + 5
$\Rightarrow\frac{\text{dy}}{\text{dx}}=1$
Now,
y = x3 + ax + b ...(1)
$\Rightarrow\frac{\text{dy}}{\text{dx}}=3\text{x}^2+\text{a}$
Slope of the tangent at (1, -6) = slope of the given line
$\Rightarrow\frac{\text{dy}}{\text{dx}}_{(1,-6)}=1$
⇒ 3 + a = 1
⇒ a = -2
On substituting a = -2, x = 1 and y = -6 in eq. (1), we get
-6 = 1 - 2 + b
⇒ b = -5
$\therefore$ a = -2 and b = -5
View full question & answer→Question 44 Marks
At what point of the curve y = x2 does the tangent make an angle of 45° with the x-axis?
AnswerSince, the point lies on the curve.
Hence, $\text{y}_1^2=\text{x}_1$
Now, $\text{y}^2=\text{x}$
$\Rightarrow2\text{y}\frac{\text{dy}}{\text{dx}}=1$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{2\text{y}}$
Slope of the tangent $=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(\text{x}_1,\text{ y}_1)}=\frac{1}{2\text{y}_1}$
Given:
$\frac{1}{2\text{y}_1}=1$
$\Rightarrow2\text{y}_1=1$
$\Rightarrow\text{y}_1=\frac{1}{2}$
Now,
$\text{x}_1=\text{y}_1^2=\Big(\frac{1}{2}\Big)^2=\frac{1}{4}$
$\therefore(\text{x}_1,\text{y}_1)=\Big(\frac{1}{4},\frac{1}{2}\Big)$
View full question & answer→Question 54 Marks
Prove that the curves xy = 4 and x2 + y2 = 8 touch each other.
AnswerGiven:
xy = 4 ...(1)
x2 + y2 = 8 ...(2)
From (1), we get
$\text{x}=\frac{4}{\text{y}}$
Substituting $\text{x}=\frac{4}{\text{y}}$ in (2), we get
$\big(\frac{4}{\text{y}}\big)^2+\text{y}^2=8$
$\Rightarrow\frac{16}{\text{y}^2}+\text{y}^2=8$
$\Rightarrow16+\text{y}^4=8\text{y}^2$
$\Rightarrow\text{y}^4-8\text{y}^2+16=0$
$\Rightarrow(\text{y}^2-4)^2=0$
$\Rightarrow\text{y}^2-4=0$
$\Rightarrow\text{y}^2=4$
$\Rightarrow\text{y}=\pm2$
Substituting $\text{y}=\pm2$ we get
$\text{x}=\pm2$
So, the given curve touches each other at two points (2, 2) and (-2, -2).
View full question & answer→Question 64 Marks
Show that the curve $\frac{\text{x}^2}{\text{a}^2+\lambda_1}+\frac{\text{y}^2}{\text{b}^2+\lambda_1}=1$ and $\frac{\text{x}^2}{\text{a}^2+\lambda_2}+\frac{\text{y}^2}{\text{b}^2+\lambda_2}=1$ intersect at right angles.
Answer we have $\frac{\text{x}^2}{\text{a}^2+\lambda_1}+\frac{\text{y}^2}{\text{b}^2+\lambda_1}=1\ ...(1)$ $\frac{\text{x}^2}{\text{a}^2+\lambda_2}+\frac{\text{y}^2}{\text{b}^2+\lambda_2}=1\ ...(2)$
Now we can find the slope of both the curve by differentiating w.r.t.x,
$\Rightarrow\frac{2\text{x}}{\text{a}^2+\lambda_1}+\frac{2\text{y}\frac{\text{dy}}{\text{dx}}}{\text{b}^2+\lambda_1}=0\text{ and }\frac{2\text{x}}{\text{a}^2+\lambda_2}+\frac{2\text{y}\frac{\text{dy}}{\text{dx}}}{\text{b}^2+\lambda_2}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{x}}{\text{y}}\times\frac{\text{b}^2+\lambda_1}{\text{a}^2+\lambda_1}\text{ and }\frac{\text{dy}}{\text{dx}}=-\frac{\text{x}}{\text{y}}\times\frac{\text{b}^2+\lambda_2}{\text{a}^2+\lambda_2}$
$\Rightarrow\text{m}_1=-\frac{\text{x}}{\text{y}}\times\frac{\text{b}^2+\lambda_1}{\text{a}^2+\lambda_1}\text{ and }\Rightarrow\text{m}_1=-\frac{\text{x}}{\text{y}}\times\frac{\text{b}^2+\lambda_2}{\text{a}^2+\lambda_2}$
Substracting (2) from (1), we get,
$\text{x}^2\Big(\frac{1}{\text{a}^2+\lambda_1}-\frac{1}{\text{a}^2+\lambda_2}\Big)+\text{y}^2\Big(\frac{1}{\text{b}^2+\lambda_1}-\frac{1}{\text{b}^2+\lambda_2}\Big)=0$
$\Rightarrow\frac{\text{x}^2}{\text{y}^2}=\frac{\lambda_2-\lambda_1}{(\text{b}^2+\lambda_1)(\text{b}^2+\lambda_2)}\times\frac{1}{\frac{\lambda_1-\lambda_2}{(\text{e}^2+\lambda_1)(\text{e}^2+\lambda_2)}}$
now,
$\text{m}_1\times\text{m}_2=\frac{\text{x}^2}{\text{y}^2}\times\frac{\text{b}^2+\lambda_1}{\text{a}^2+\lambda_1}\times\frac{\text{b}^2+\lambda_2}{\text{a}^2+\lambda_2}$
$=\frac{\lambda_2-\lambda_1}{(\text{b}^2+\lambda_1)(\text{b}^2+\lambda_2)}\times\frac{(\text{a}^2+\lambda_1)(\text{a}^2+\lambda_2)}{\lambda_1-\lambda_2}\times\frac{\text{b}^2+\lambda_1}{\text{a}^2+\lambda_1}\times\frac{\text{b}^2+\lambda_1}{\text{a}^2+\lambda_1}$
$=-1$
Hence, (1) and (2) cuts orthogonally.
View full question & answer→Question 74 Marks
Find the equations of the tangent and the normal to the following curves at the indicated points.
$\text{x}^{\frac{2}{3}}+\text{y}^{\frac{2}{3}}=2\text{ at }(1,1)$
Answer$\text{x}^{\frac{2}{3}}+\text{y}^{\frac{2}{3}}=2$ Differentiating both sides w.r.t.x,
$\frac{2}{3}\text{x}^{\frac{-1}{3}}+\frac{2}{3}\text{y}^{\frac{-1}{3}}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-\text{x}^{\frac{-1}{3}}}{\text{y}^{\frac{-1}{3}}}=\frac{-\text{y}^{\frac{1}{3}}}{\text{x}^{\frac{1}{3}}}$
Slope of tangent,
$\text{m}=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(1,1)}=\frac{-1}{1}=-1$ Given
$(\text{x}_1,\text{y}_1)=(1,1)$ Eequation of tangent is,
$\text{y}-\text{y}_1=\text{m}(\text{x}-\text{x}_1)$
$\Rightarrow\text{y}-1=-1(\text{x}-1)$
$\Rightarrow\text{y}-1=-\text{x}+1$
$\Rightarrow\text{x}+\text{y}-2=0$
Equation of normal is,
$\text{y}-\text{y}_1=\frac{-1}{\text{m}}(\text{x}-\text{x}_1)$ $\Rightarrow\text{y}-1=1(\text{x}-1)$
$\Rightarrow\text{y}-1=\text{x}-1$
$\Rightarrow\text{y}-\text{x}=0$
View full question & answer→Question 84 Marks
Show that the following curves intersect orthogonally at the indicated points:
x2 = 4y and 4y + x2 = 8 at (2, 1)
Answerx2 = 4y ...(1)
4y + x2 = 8...(2)
Given point is (2, 1)
Differentiating (1) w..r.t.x,
$2\text{x}=4\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{2}$
$\Rightarrow\text{m}_1=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(2, 1)}=\frac{2}{2}=1$
Differentiating (2) w..r.t.x,
$4\frac{\text{dy}}{\text{dx}}+2\text{x}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-\text{x}}{2}$
$\Rightarrow\text{m}_2=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(2,1)}=\frac{-2}{2}=-1$
Since, $\text{m}_1\times\text{m}_2=-1$
View full question & answer→Question 94 Marks
Find the equations of the tangent and the normal to the following curves at the indicated points.
$\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}=1\text{ at }(\text{x}_0,\text{y}_0)$
Answer$\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}=1$
Differentiating both sides w.r.t.x,
$\frac{2\text{x}}{\text{a}^2}-\frac{2\text{y}}{\text{b}^2}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{2\text{y}}{\text{b}^2}\frac{\text{dy}}{\text{dx}}=\frac{2\text{x}}{\text{a}^2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{xb}^2}{\text{ya}^2}$
Slope of tangent, $\text{m}=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(\text{x}_0,\text{y}_0)}=\frac{\text{x}_0\text{b}^2}{\text{y}_0\text{a}^2}$
Equation of tangent is,
$\text{y}-\text{y}_1=\text{m}(\text{x}-\text{x}_1)$
$\Rightarrow\text{y}-\text{y}_0=\frac{\text{x}_0\text{b}_0}{\text{y}_0\text{a}^2}(\text{x}-\text{x}_0)$
$\Rightarrow\text{yy}_0\text{a}^2-\text{y}_0{^2}=\text{xx}_0\text{b}^2-\text{x}_0{^2}\text{b}^2$
$\text{xx}_0\text{b}^2-\text{yy}_0\text{a}^2=\text{x}_0{^2}\text{b}^2-\text{y}_0{^2}\text{a}^2\ ...(1)$
Since $(\text{x}_0,\text{y}_0)$ lies on the given curve,
$\Rightarrow\frac{\text{x}_0{^2}}{\text{a}^2}-\frac{\text{y}_0{^2}}{\text{b}^2}=1$
$\Rightarrow\text{x}_0{^2}\text{b}^2-\text{y}_0{^2}=\text{a}^2\text{b}^2$
Substituting this in (1), we get
$\Rightarrow\text{xx}_0\text{b}^2-\text{yy}_0\text{a}^2=\text{a}^2\text{b}^2$
Dividing this by $\text{a}^2\text{b}^2$
$\frac{\text{xx}_0}{\text{a}^2}-\frac{\text{yy}_0}{\text{b}^2}=1$
Equation of normal is,
$\text{y}-\text{y}_1=\text{m}(\text{x}-\text{x}_1)$
$\Rightarrow\text{y}-\text{y}_0=\frac{-\text{y}_0\text{a}^2}{\text{x}_0\text{b}^2}(\text{x}-\text{x}_0)$
$\Rightarrow\text{yx}_0\text{b}^2-\text{x}_0\text{y}_0\text{b}^2=-\text{xy}_0\text{a}^2+\text{x}_0\text{y}_0\text{a}^2$
$\Rightarrow\text{xy}_0\text{a}^2+\text{yx}_0\text{b}^2=\text{x}_0\text{y}_0\text{a}^2+\text{x}_0\text{y}_0\text{b}^2$
$\Rightarrow\text{xy}_0\text{a}^2+\text{yx}_0\text{b}^2=\text{x}_0\text{y}_0(\text{a}^2+\text{b}^2)$
Dividing by $\text{x}_0\text{y}_0$
$\frac{\text{a}^2\text{x}}{\text{x}_0}+\frac{\text{b}^2\text{y}}{\text{y}_0}=\text{a}^2+\text{b}^2$
View full question & answer→Question 104 Marks
Find the angle of intersecting of the following curves:
$\text{y}^2=\text{x}\text{ and }\text{x}^2=\text{y}$
AnswerGiven curves are, y2 = x ...(1) x2 = y...(2) From these two equations, we get $(\text{x}^2)^2=\text{x}$ $\Rightarrow\text{x}^4-\text{x}=0$ $\Rightarrow\text{x}(\text{x}^3-1)=0$ $\Rightarrow\text{x}=0\text{ or }\text{x}=1$ Substituting the value of x in (2) we get, y = 0 or y = 1 $\therefore(\text{x},\text{y})=(0,0)\text{ or }(1,1)$ Differentiating (1) w.r.t.x, $2\text{y}\frac{\text{dy}}{\text{dx}}=1$ $\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{2\text{y}}\ ...(3)$ Differentiating (2) w.r.t.x, $2\text{x}=\frac{\text{dy}}{\text{dx}}\ ...(4)$ Case 1: (x, y) = (0, 0) The tangent to curve is parallel to x-axis Hence, the angle between the tangent to two curve at (0, 0) is a right angle. $\therefore\theta=\frac{\pi}{2}$ Case 2:
(x, y) = (1, 1) From (3) we have, $\text{m}_1=\frac{1}{2}$ From (4) we have, $\text{m}_2=2(1)=2$ Now, $\tan\theta=\big|\frac{\text{m}_1-\text{m}_2}{1+\text{m}_1\text{m}_2}\big|=\Big|\frac{\frac{1}{2}-2}{1+\frac{1}{2}\times2}\Big|=\frac{3}{4}$ $\Rightarrow\theta=\tan^{-1}\big(\frac{3}{4}\big)$ View full question & answer→Question 114 Marks
Find the condition for the following set of curves to intersect orthogonally
$\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}=1\text{ and }\text{xy}=\text{c}^2$
Answerwe have,
$\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}=1\ ...(1)$
$\text{xy}=\text{c}^2...(2)$
Slope if (1)
$\frac{2\text{x}}{\text{a}^2}-\frac{2\text{y}}{\text{b}^2}\frac{\text{dy}}{\text{dx}}=0$
$\therefore\text{m}_1=\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\text{y}}\times\frac{\text{b}^2}{\text{a}^2}$
Slope if (2)
$\text{y}+\text{x}\frac{\text{dy}}{\text{dx}}=0$
$\therefore\text{m}_2=\frac{\text{dy}}{\text{dx}}=\frac{-\text{y}}{\text{x}}$
(1) and (2) cuts orthogonally
$\therefore\text{m}_1\times\text{m}_2=-1$
$\Rightarrow\frac{\text{x}}{\text{y}}\times\frac{-\text{y}}{\text{x}}\times\frac{\text{a}^2}{\text{b}^2}=-1$
$\Rightarrow\text{a}^2=\text{b}^2$
View full question & answer→Question 124 Marks
Find the slopes of the tangent and the normal to the following curves at the indicated points:
$\text{x}=\text{a}(\theta-\sin\theta),\text{y}=\text{a}(1-\cos\theta)\ \text{at}\ \theta=\frac{-\pi}{2}$
AnswerWe know that the slope of the tangent to the curve y = f(x) is
$\frac{\text{dy}}{\text{dx}}=\text{f}'(\text{x})\ ...(1)$
And the slope of the normal is
$\frac{-1}{\frac{\text{dy}}{\text{dx}}}=\frac{-1}{\text{f}'(\text{x})}\ ...(2)$
$\text{x}=\text{a}(\theta-\sin\theta),\text{y}=\text{a}(1+\cos\theta)$
Now,
$\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}=\frac{-\text{a}\sin\theta}{\text{a}(1-\cos\theta)}$
$\therefore$ slope of tangent of $\theta=-\frac{\pi}{2}$
$\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\theta-\frac{\pi}{2}}=\frac{-\text{a}\sin\Big(-\frac{\pi}{2}\Big)}{\text{a}\Big(1-\cos\big(-\frac{\pi}{2}\big)\Big)}$
$=\frac{\text{a}}{\text{a}(1-0)}=1$
Also, the slope of normal is
$\frac{-1}{\frac{\text{dy}}{\text{dx}}}=\frac{-1}{\text{f}'(\text{x})}=-1$
View full question & answer→Question 134 Marks
At what points on the curve y = 2x2 - x + 1 is the tangent parallel to the line y = 3x + 4?
AnswerLet (x1, y1) be the required point.
The slope of line y = 3x + 4 is 3.
Since, the point lies on the curve.
Hence, $\text{y}_1=2\text{x}_1^2-\text{x}_1+1$
Now, $\text{y}=2\text{x}^2-\text{x}+1$
$\frac{\text{dy}}{\text{dx}}=4\text{x}-1$
Now,
Slope of thge tangent at $(\text{x}_1,\text{y}_1)=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(\text{x}_1,\text{ y}_1)}=4\text{x}_1-1$
Slope of thge tangent at $(\text{x}_1,\text{ y}_1)$ = slope of the given line [Given]
$\therefore4\text{x}_1-1=3$
$\Rightarrow4\text{x}_1=4$
$\Rightarrow\text{x}_1=1$
And
$\text{y}_1=2\text{x}_1^2-\text{x}_1+1=2-1+1=2$
Thus, the required point is (1, 2).
View full question & answer→Question 144 Marks
Show that the following curves intersect orthogonally at the indicated points:
x2 = y and x3 + 6y = 7 at (1, 1)
Answerwe have,
x2 = y ...(1)
x3 + 6y = 7 ...(2)
Slope of (i)
$2\text{x}=\frac{\text{dy}}{\text{dx}}$
$\therefore\text{m}_1\Big(\frac{\text{dy}}{\text{dx}}\Big)_\text{p}=2$
Slope of (ii)
$3\text{x}^2+6\frac{\text{dy}}{\text{dx}}=0$
$\therefore\text{m}_2=\Big(\frac{\text{dy}}{\text{dx}}\Big)_\text{p}=\Big(-\frac{\text{x}^2}{2}\Big)_\text{p}=\frac{-1}{2}$
$\therefore\text{m}_1\times\text{m}_2=2\times\frac{-1}{2}=-1$
View full question & answer→Question 154 Marks
Find the equation of the tangent to the curve $\text{x}=\sin3\text{t},\text{y}=\cos2\text{t}\text{ at }\text{t}=\frac{\pi}{4}$
Answer$\text{x}=\sin3\text{t},\text{y}=\cos2\text{t}$
$\frac{\text{dx}}{\text{dt}}=3\cos3\text{t}\ \text{and}\ \frac{\text{dy}}{\text{dt}}=-2\sin2\text{t}$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{-2\sin2\text{t}}{3\cos3\text{t}}$
Slope of tangent, $\text{m}=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{t}=\frac{\pi}{4}}=-\frac{-2\sin\big(\frac{\pi}{2}\big)}{3\cos\big(\frac{3\text{x}}{4}\big)}=\frac{-2}{\frac{-3}{\sqrt{2}}}=\frac{2\sqrt{2}}{3}$
$\text{x}_1=\sin\big(3\times\frac{\pi}{4}\big)=\frac{1}{\sqrt{2}}\text{ and }\text{y}_1=\cos\big(2\times\frac{\pi}{4}\big)=0$
so, $(\text{x}_1,\text{y}_1)=\big(\frac{1}{\sqrt{2}},2\big)$
Equation of tangent is,
$\text{y}-\text{y}_1=\text{m}(\text{x}-\text{x}_1)$
$\Rightarrow\text{y}-0=\frac{2\sqrt{2}}{3}\big(\text{x}-\frac{1}{\sqrt{2}}\big)$
$\Rightarrow3\text{y}=2\sqrt{2}\text{x}-2$
$\Rightarrow2\sqrt{2}-3\text{y}-2=0$
View full question & answer→Question 164 Marks
Show that the following set of curves intersect orthogonally.
x3 - 3xy2 = -2 and 3x2y - y3 = 2
Answerwe know that two curves interesects orthogonally if
$\text{m}_1\times\text{m}_2=-1\ ...(\text{A})$
Where m1 and m2 are the slopes of two curves
$\text{x}^3-3\text{xy}^2=-2\ ...(1)$
$3\text{x}^2\text{y}-\text{y}^3=2\ ...(2)$
Point of intersection of (1) and (2)
(1) + (2)
$\Rightarrow\text{x}^3-3\text{xy}^2+3\text{x}^2\text{y}-\text{y}^3=0$
$\Rightarrow(\text{x}-\text{y})^3=0$
$\Rightarrow\text{x}=\text{y}$
From (1)
$\text{x}^3-3\text{x}^2=-2$
$\Rightarrow-2\text{x}^3=-2$
$\Rightarrow\text{x}=1$
$\therefore$ P = (1, 1) is the point of intersection
now,
slope of (1)
$3\text{x}^2-3\text{y}^2-6\text{xy}\frac{\text{dy}}{\text{dx}}=0$
$\therefore\text{m}_1=\frac{\text{dy}}{\text{dx}}=\frac{3(\text{x}^2-\text{y}^2)}{6\text{xy}}$
slope of (2)
$6\text{xy}+3\text{x}^2\frac{\text{dy}}{\text{dx}}-3\text{y}^2\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\text{m}_2=\frac{\text{dy}}{\text{dx}}=\frac{-6\text{xy}}{3(\text{x}^2-\text{y}^2)}$
$\therefore\text{m}_1\times\text{m}_2=\frac{(\text{x}^2-\text{y}^2)}{2\text{xy}}\times\frac{-2\text{xy}}{(\text{x}^2-\text{y}^2)}=-1$
View full question & answer→Question 174 Marks
Find the equation of the normal to the curve x2 + 2y2 - 4x - 6y + 8 = 0 at the point whose abscissa is 2.
AnswerThe given equation of curve is
x2 + 2y2 - 4x - 6y + 8 = 0
Differentiating with respect to x, we get
$2\text{x}+4\text{y}\frac{\text{dy}}{\text{dx}}-4-6\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}[4\text{y}-6]=-2\text{x}$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{2-\text{x}}{2\text{y}-3}$
Now,
From (i) at x = 2
4 + 2y2 - 8 - 6y + 8 = 0
⇒ 2y2 - 6y + 4 = 0
⇒ y2 - 3y + 2 = 0
⇒ (y - 2) (y - 1) = 0
⇒ y = 2, 1
Thus,
Slope $\text{m}_1=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(2,2)}=0$
$\text{m}_2=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(2,1)}=0$
Thus, the equation of normal is
$(\text{y}-\text{y}_1)=\frac{-1}{0}(\text{x}-2)$
$\Rightarrow\text{x}=2$
View full question & answer→Question 184 Marks
Find the angle of intersecting of the following curves:
$\text{x}^2+\text{y}^2-4\text{x}-1=0\text{ and }\text{x}^2+\text{y}^2-2\text{y}-9=0$
AnswerGiven curve are,
$\text{x}^2+\text{y}^2-4\text{x}-1=0\ ...(1)$
$\text{x}^2+\text{y}^2-2\text{y}-9=0\ ...(2)$
From (3) we get
x2 + y2 = 4x + 1
Substituting this in (2),
4x + 1 - 2y - 9 = 0
⇒ 4x - 2y = 8
⇒ 2x - y = 4
⇒ y = 2x - 4 ...(3)
Substituting this in (1),
x2 + (2x - 4)2 - 4x - 1 = 0
⇒ x2 + 4x2 + 16 - 16x - 4x - 1 = 0
⇒ 5x2 - 20x + 15 = 0
⇒ x2 - x + 3 = 0
⇒ (x - 3) (x - 1) = 0
⇒ x = 3 or x = 1
substituting the value of x in (3), we get,
y = 2 or y = -2
$\therefore$ (x, y) = (3, 2), (1, -2)
Differentiating (1) w.r.t.x,
$2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}-4=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{4-2\text{x}}{2\text{y}}=\frac{2-\text{x}}{\text{y}}\ ...(4)$
Differentiating (2) w.r.t.x,
$2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}-2\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}(2\text {y}-2)=-2\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2\text{x}}{2-2\text{y}}=\frac{\text{x}}{1-\text{y}}\ ...(5)$
Case 1:
(x, y) = (3, 2)
From (4), we get, $\text{m}_1=\frac{2-3}{2}=\frac{-1}{2}$
From (5), we get, $\text{m}_2=\frac{3}{1-2}=-3$
Now,
$\tan\theta=\Big|\frac{\text{m}_1-\text{m}_2}{1+\text{m}_1\text{m}_2}\Big|=\Bigg|\frac{\frac{-1}{2}+3}{1+\frac{3}{2}}\Bigg|=1$
$\Rightarrow\theta=\tan^{-1}(1)=\frac{\pi}{4}$
Case 2:
(x, y) = (1, -2)
From (4), we get, $\text{m}_1=\frac{2-1}{-2}=\frac{-1}{2}$
From (5), we get, $\text{m}_2=\frac{1}{1+2}=\frac{1}{3}$
Now,
$\tan\theta=\Big|\frac{\text{m}_1-\text{m}_2}{1+\text{m}_1\text{m}_2}\Big|=\Bigg|\frac{\frac{-1}{2}-\frac{1}{3}}{1-\frac{1}{6}}\Bigg|=1$
$\Rightarrow\theta=\tan^{-1}(1)=\frac{\pi}{4}$
View full question & answer→Question 194 Marks
Find the equations of all lines having slope 2 and that are tangent to the curve $\text{y}=\frac{1}{\text{x}=3},\text{x}\neq3.$
AnswerSlope of given tangent = 2
Let $(\text{x}_1,\text{y}_1)$ be the point where the tangent is drawn to this curve.
Since, the point lies on the curve.
Hence, $\text{y}_1=\frac{1}{\text{x}_1-3}$
Now, $\text{y}=\frac{1}{\text{x}-3}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-1}{(\text{x}-3)^2}$
Slope of tangent $=\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{-1}{{(\text{x}_1-3})^2}$
Given that
Slope of tangent = 2
$\Rightarrow\frac{-1}{(\text{x}_1-3)^2}=2$
$\Rightarrow(\text{x}_1-3)^2=-2$
$\Rightarrow\text{x}_1-3=\sqrt{2},$
Which does not exiast because 2 is negative.
So, there does not exist any such tangent.
View full question & answer→Question 204 Marks
Find the equations of the tangent and the normal to the following curves at the indicated points.
$\text{x}=3\cos\theta-\cos^3\theta,\text{y}=3\sin\theta-\sin^3\theta$
Answer $\text{x}=3\cos\theta-\cos^3\theta,\text{y}=3\sin\theta-\sin^3\theta$ $\Rightarrow\frac{\text{dx}}{\text{d}\theta}=-3\sin\theta+3\cos^2\theta\sin\theta$
$\text{and }\frac{\text{dy}}{\text{d}\theta}=3\cos\theta-3\sin\theta\cos\theta$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}=\frac{3\cos\theta-3\sin^2\theta\cos\theta}{-3\sin\theta+3\cos^2\theta\sin\theta}$
$=\frac{\cos\theta(1-\sin^2\theta)}{-\sin\theta(1-\cos^2\theta)}=\frac{\cos^3\theta}{-\sin^\theta}=\tan^3\theta$
So, equation o the tangent at $\theta$ is
$\text{y}-3\sin\theta+\sin^3\theta=-\tan^3\theta(\text{x}-3\cos\theta+\cos^\theta)$
$\Rightarrow4(\text{y}\cos^3\theta-\text{x}\sin^3\theta)=3\sin4\theta$
So, equation o the normal at $\theta$ is
$\text{y}-3\sin\theta+\sin^3\theta=-\frac{1}{\tan^3\theta}(\text{x}-3\cos\theta+\cos^\theta)$
$\Rightarrow\text{y}\cos^3\theta-\text{x}\cos^3\theta=3\sin^4\theta-\sin^6\theta-3\cos^4\theta+\cos^6\theta$
$\Rightarrow\text{y}\sin^3\theta-\text{x}\cos^3\theta=3\sin^4\theta-\sin^6\theta-3\cos^4\theta+\cos^6\theta$
View full question & answer→Question 214 Marks
Find the point on the curve $\frac{\text{x}^2}{9}+\frac{\text{y}^2}{16}=1$ at which the tangent are:
- parallel to x-axis
- paralle to y- axis
AnswerThe equation of the given curve is $\frac{\text{x}^2}{9}+\frac{\text{y}^2}{16}=1$
On differentiating both sides with respect to x, we have:
$\frac{2\text{x}}{9}+\frac{2\text{y}}{16}.\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-16\text{x}}{9\text{y}}$
The tangent is parallel to the x axis if the slope of the tangent is i.e, 0 $\frac{-16\text{x}}{9\text{y}}=0,$ which is possible if x = 0.
Then, $\frac{-1}{\Big(\frac{-16\text{x}}{9\text{y}}\Big)}=\frac{9\text{y}}{16\text{x}}=0\Rightarrow\text{y}=0.$
Then, $\frac{\text{x}^2}{9}+\frac{\text{y}^2}{16}=1$ for y = 0.
$\Rightarrow\text{x}=\pm 3$
Hence, the point at which the tangents are parallel to the y axis are,
(3, 0) and (-3, 0)
View full question & answer→Question 224 Marks
Find the equation of the tangent to the curve $\text{x}=\theta+\sin\theta,\text{y}+\cos\theta\text{ at }\theta=\frac{\pi}{4}.$
Answer$\text{x}=\theta+\sin\theta,\text{y}=1+\cos\theta$
$\frac{\text{dx}}{\text{d}\theta}=1+\cos\theta\text{ and }\frac{\text{dy}}{\text{d}\theta}=-\sin\theta$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}=\frac{-\sin\theta}{1+\cos\theta}$
Slope of tangent $=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\theta=\frac{\pi}{4}}=\frac{-\sin\frac{\pi}{4}}{1+\cos\frac{\pi}{4}}=\frac{\frac{-1}{\sqrt{2}}}{1+\frac{1}{\sqrt{2}}}=\frac{-1}{\sqrt{2}+1}\times\frac{\sqrt{2}-1}{\sqrt{2}-1}=1-\sqrt{2}$
$(\text{x}_1,\text{y}_1)=\Big(\frac{\pi}{4}+\sin\frac{\pi}{4},1+\cos\frac{\pi}{4}\Big)=\Big(\frac{\pi}{4}+\frac{1}{\sqrt{2}},1+\frac{1}{\sqrt{2}}\Big)$
Equation of tangent is,
$\text{y}-\text{y}_1=\text{m}(\text{x}-\text{x}_1)$
$\Rightarrow\text{y}-\Big(1+\frac{1}{\sqrt{2}}\Big)=(1-\sqrt{2})\Big[\text{x}-\text{x}-\Big(\frac{\pi}{4}+\frac{1}{\sqrt{2}}\Big)\Big]$
$\Rightarrow\text{y}-1-\frac{1}{\sqrt{2}}=(1-\sqrt{2})\Big[\text{x}-\frac{\pi}{4}-\frac{1}{\sqrt{2}}\Big]$
View full question & answer→Question 234 Marks
Find a point on the curve y = x3 - 3x where the tangent is parallel to the chord joining (1, -2) and (2, 2).
AnswerSlope of the chord $=\frac{\text{y}_2-\text{y}_1}{\text{x}_2-\text{x}_1}=\frac{2+2}{2-1}=4$ $\text{y}=\text{x}^3=3\text{x}$ $\Rightarrow\frac{\text{dy}}{\text{dx}}=3\text{x}^2-3\ ...(1)$ Slope of the tangent $=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(\text{x}_1,\text{y}_1)}=3\text{x}_1^2-3$ It is given that the tangent and the chord are parallel. $\therefore$ slope of the tangent = slope of the chord $\Rightarrow3\text{x}_1^2-3=4$ $\Rightarrow3\text{x}_1^2=7$ $\Rightarrow\text{x}_1^2=\frac{7}{3}$ $\Rightarrow\text{x}_1=\pm\sqrt{\frac{7}{3}}=\sqrt{\frac{7}{3}}\text{or}-\sqrt{\frac{7}{3}}$ Case 1
When $\text{x}_1=\sqrt{\frac{7}{3}}$ On substituting this in eq. (1), we get $\text{y}_1=\Big(\sqrt{\frac{7}{3}}\Big)^3-3\Big(\sqrt{\frac{7}{3}}\Big)=\frac{7}{3}\sqrt{\frac{7}{3}}-3\sqrt{\frac{7}{3}}=\frac{-2}{3}\sqrt{\frac{7}{3}}$ $(\text{x}_1,\text{y}_1)=\Big(\sqrt{\frac{7}{3}},\frac{-2}{3}\sqrt{\frac{7}{3}}\Big)$ Case 2
when $\text{x}_1=-\sqrt{\frac{7}{3}}$ View full question & answer→Question 244 Marks
Find the equations of the tangent and the normal to the following curves at the indicated points.
$\text{x}=\theta+\sin\theta,\text{y}=1+\cos\theta\text{ at }\theta=\frac{\pi}{2}$
AnswerWe know taht the equation of tangent and npormal to any curve at the point (x1, y1) is
$\text{y}-\text{y}_1=\text{m}(\text{x}-\text{x}_1)\ ...(1)\ \text{Tangent}$
$\text{y}-\text{y}_1=\frac{-1}{\text{m}}(\text{x}-\text{x}_1)\ ...(2) \ \text{Normal}$
Where m is slope.
$\text{x}=\theta+\sin\theta,\text{y}=1+\cos\theta,\theta=\frac{\pi}{2}$
$\therefore\text{p}=\Big[\big(\frac{\pi}{2}+1\big),1\Big]$
And $\frac{\text{dx}}{\text{d}\theta}=1+\cos\theta,\frac{\text{dy}}{\text{d}\theta}=-\sin\theta$
$\therefore$ Slope $\text{m}=\Big(\frac{\text{dy}}{\text{dx}}\Big)_\text{p}=\Bigg(\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}\Bigg)=\frac{-1}{+1}=-1$
Equation of tangent from (A)
$(\text{y}-1)=-1\Big(\text{x}-\big(\frac{\pi}{2}+1\big)\Big)$
$\Rightarrow\text{x}+\text{y}=\frac{\pi}{2}+1+1$
$\Rightarrow2(\text{x}+\text{y})=\pi+4$
From (B)
Equation of normal is
$(\text{y}-1)=1\Big(\text{x}-\big(\frac{\pi}{2}+1\big)\Big)$
$\Rightarrow2(\text{x}-\text{y})=\pi$
View full question & answer→Question 254 Marks
Find the points on the curve y = x3 - 2x2 - 2x at which the tangent lines are parallel to the line y = 2x - 3.
AnswerThe given equations are
y = x3 - 2x2 - 2x ...(i)
y = 2x - 3 ...(ii)
Slope to the tangents of (i) and (ii) are
$\frac{\text{dy}}{\text{dx}}=3\text{x}^2-4\text{x}-2\ ...(\text{iii)}$
and $\frac{\text{dy}}{\text{dx}}=2\ ...(\text{iv})$
According to the question slope to (i) and (ii) are parallel, so
3x2 - 4x - 2 = 2
⇒ 3x2 - 4x - 4 = 0
⇒ 3x2 - 6x - 2x - 4 = 0
⇒ 3x(x - 2) + 2(x - 2) = 0
$\Rightarrow\text{x}=\frac{-2}{3}\text{ or}\ 2$
From (i)
$\text{y}=\frac{4}{27}\ \text{or}-4$
Thus, the point are
$\Big(\frac{-2}{3},\frac{4}{27}\Big)\ \text{and}\ (2,-4)$
View full question & answer→Question 264 Marks
Find the angle of intersecting of the following curves:
$2\text{y}^2=\text{x}^3\text{ and }\text {y}^2=32\text{x}$
AnswerGiven curve are,
2y2 = x3 ...(1)
y2 = 32x ...(2)
From the two equations we get
2(32x) = x3
⇒ 64x = x3
⇒ x(x2 - 64) = 0
⇒ x = 0, 8, -8
Substituting the value of x in (2) we get,
y1 = 0, 16, -16
$\therefore$ (x1, y1) = (0, 0), (8, 16) or (8, -16)
Differentiating (2) w.r.t.x,
$4\text{y}\frac{\text{dy}}{\text{dx}}=3\text{x}^2$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{3\text{x}^2}{4\text{y}}\ ...(3)$
Differentiating (3) w.r.t.x,
$2\text{y}\frac{\text{dy}}{\text{dx}}=32$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{16}{\text{y}}\ ...(4)$
Case 1:
(x, y) = (0, 0)
From (3) we have, $\text{m}_1=\frac{0}{0}$
$\therefore$ we cannot determine $\theta$ in this case.
Case 2:
(x, y) = (8, 16)
From (3) we have, $\text{m}_1=\frac{192}{64}=3$
From (4) we have, $\text{m}_2=\frac{16}{-16}=-1$
Now, $\tan\theta=\Big|\frac{\text{m}_1-\text{m}_2}{1+\text{m}_1\text{m}_2}\Big|=\Big|\frac{-3+1}{1+3}\Big|=\frac{2}{4}=\frac{1}{2}$
$\Rightarrow\theta=\tan^{-1}\big(\frac{1}{2}\big)$
View full question & answer→Question 274 Marks
Find the equations of the tangent and the normal to the following curves at the indicated points.
$\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1\text{ at }(\text{a}\cos\theta,\text{b}\sin\theta)$
Answer$\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1$
Differentiating both sides w.r.t.x,
$\Rightarrow\frac{2\text{x}}{\text{a}^2}+\frac{2\text{y}}{\text{b}^2}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{2\text{y}}{\text{b}^2}\frac{\text{dy}}{\text{dx}}=\frac{-2\text{x}}{\text{a}^2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-\text{xb}^2}{\text{ya}^2}$
Slope of tangent, $\text{m}=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(\text{a}\cos\theta,\text{b}\sin\theta)}=\frac{-\text{a}\cos\theta(\text{b})^2}{\text{b}\sin\theta(\text{a}^2)}=\frac{-\text{b}\cos\theta}{\text{a}\sin\theta}$
Given $(\text{x}_1,\text{y}_1)=(\text{a}\cos\theta,\text{b}\sin\theta)$
Equation of tangent is,
$\text{y}-\text{y}_1=\text{m}(\text{x}-\text{x}_1)$
$\Rightarrow\text{y}-\text{b}\sin\theta=\frac{-\text{b}\cos\theta}{\text{a}\sin\theta}(\text{x}-\text{a}\cos\theta)$
$\Rightarrow\text{ay}\sin\theta-\text{ab}\sin^2\theta=-\text{bx}\cos\theta+\text{ab}\cos^2\theta$
$\Rightarrow\text{bx}\cos\theta+\text{ay}\sin\theta=\text{ab}$
Dividing by ab,
$\Rightarrow\frac{\text{x}}{\text{a}}\cos\theta+\frac{\text{y}}{\text{b}}\sin\theta=1$
Equation of normal is,
$\text{y}-\text{y}_1=\frac{-1}{\text{m}}(\text{x}-\text{x}_1)$
$\Rightarrow\text{y}-\text{b}\sin\theta=\frac{\text{a}\sin\theta}{\text{b}\cos\theta}(\text{x}-\text{A}\cos\theta)$
$\Rightarrow\text{by}\cos\theta-\text{b}^2\sin\theta\cos\theta=\text{ax}\sin\theta-\text{a}^2\sin\theta\cos\theta$
$\Rightarrow\text{ax}\sin\theta-\text{by}\cos\theta=(\text{a}^2-\text{b}^2)\sin\theta\cos\theta$
Dividing by $\sin\theta\cos\theta$
$\text{ax}\sec\theta-\text{by }\text{cosec}\theta=(\text{a}^2-\text{b}^2)$
View full question & answer→Question 284 Marks
Show that the following set of curves intersect orthogonally.
y = x3 and 6y = 7 - x2
Answery = x3 ...(1)
6y = 7 - x2 ...(2)
From (1) and (2) we get,
6x3 = 7 - x2
⇒ 6x3 - x2 - 7 = 0
x = 1 satisfies this.
Dividing this by x - 1, we get
6x2 + 7x + 7 = 0
Discriminant = 72 - 4 (6) (7) = -119 < 0.
So, (x, y) = (1, 1)
Differentiating (1) w.r.t.x,
$\frac{\text{dy}}{\text{dx}}=3\text{x}^2$
$\Rightarrow\text{m}_1=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(1,1)}=3$
Differentiating (2) w.r.t. x,
$6\frac{\text{dx}}{\text{dx}}=-2\text{x}$
$\Rightarrow\frac{\text{dx}}{\text{dx}}=\frac{-2\text{x}}{6}=\frac{-\text{x}}{3}$
$\Rightarrow\text{m}_2=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(1,1)}=\frac{-1}{3}$
Now, $\text{m}_1\times\text{m}_2=3\times\frac{-1}{3}$
$\Rightarrow\text{m}_1\times\text{m}_2=-1$
Since,
$\Rightarrow\text{m}_1\times\text{m}_2=1$
So, the curve intersect orthogonally.
View full question & answer→Question 294 Marks
Find the equation of the tangent line to the curve y = x2 + 4x - 16 which is parallel to the line 3x - y + 1 = 0.
AnswerThe given equation are,
y = x2 + 4x - 16
3x - y + 1 = 0
Slope m1 if (i)
$\text{m}_1=\frac{\text{dy}}{\text{dx}}=2\text{x}$
Slope m2 if (ii)
m2 = 3
As per question
$\text{m}_1=\text{m}_2$
$\Rightarrow2\text{x}+4=3$
$\Rightarrow\text{x}=\frac{-1}{3}$
From (i)
$\text{y}=\frac{1}{4}-2-16=-\frac{71}{4}$
$\therefore\text{P}=\Big(\frac{-1}{2},\frac{-71}{4}\Big)$
Thus, the equation of tangent
$\Big(\text{y}+\frac{71}{4}\Big)=3\Big(\text{x}+\frac{1}{2}\Big)$
$\Rightarrow3\text{x}-\text{y}=\frac{71}{4}-\frac{3}{2}$
$\Rightarrow3\text{x}-\text{y}=\frac{65}{4}$
$\Rightarrow12\text{x}-4\text{y}-65=0$
View full question & answer→Question 304 Marks
Prove that the curves y2 = 4x and x2 + y2 - 6x + 1 = 0 touch each other at the point (1, 2).
AnswerGiven:
Curves y2 = 4x ...(1)
And x2 + y2 - 6x + 1 = 0 ...(2)
$\therefore$ The point of intersection of two curves is (1, 2)
First curve is y2 = 4x
Differentiating above w.r.t. x,
$\Rightarrow2\text{y}.\frac{\text{dy}}{\text{dx}}=4$
$\Rightarrow\text{y}.\frac{\text{dy}}{\text{dx}}=2$
$\Rightarrow\text{m}_1=\frac{2}{\text{y}}\dots\text{(3)}$
Second curve is x2 + y2 - 6x + 1 = 0
$\Rightarrow\text{2x}+\text{2y}.\frac{\text{dy}}{\text{dx}}-6-0=0$
$\Rightarrow\text{x}+\text{y}.\frac{\text{dy}}{\text{dx}}-3=0$
$\Rightarrow\text{y}.\frac{\text{dy}}{\text{dx}}=3-\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{3-\text{x}}{\text{y}}\dots(4)$
At (1, 2), we have,
$\text{m}_1=\frac{2}{\text{y}}$
$\Rightarrow\frac22$
$\text{m}_1=1$
At (1, 2), we have,
$\Rightarrow\text{m}_2=\frac{3-\text{x}}{\text{y}}$
$\Rightarrow\frac{3-1}{2}$
$\Rightarrow\text{m}_2=1$
Clearly, m1 = m2 = 1 at (1, 2)
So, given curve touch each other at (1, 2)
View full question & answer→Question 314 Marks
Find the equation of the tangents to the curve 3x2 - y2 = 8, which passes through the point $\big(\frac{4}{3},0\big)$
AnswerConsider the equation of the curve
3x2 - y2 = 8
Differentiating the above expression w.r.t. x,
$6\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}=0$
Slope, $\text{m}=\frac{\text{dy}}{\text{dx}}$
$\Rightarrow6\text{x}-2\text{ym}=0$
$\Rightarrow2\text{ym}=6\text{x}$
$\Rightarrow\text{m}=\frac{6\text{x}}{2\text{y}}$
$\Rightarrow\text{m}=\frac{3\text{x}}{\text{y}}$
Let P $(\text{x}_1,\text{y}_1)$ be any point on the curve.
Thus, we have
$3\text{x}_1^2-\text{y}_1^2=8$
$\Rightarrow\text{y}_1^2=3\text{x}_1^2-8\ ...(1)$
The equation of the tangent to the curve at the point P $(\text{x}_1,\text{ y}_1)$ is
$\text{y}-\text{y}_1=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{P}[\text{x}_1,\text{y}_1]}(\ \text{x}-\text{x}_1)$
Therefore, the equation of tangent is
$\text{y}-\text{y}_1=\Big(\frac{3\text{x}_1}{\text{y}_1}\Big)(\text{x}-\text{x}_1)$
This line passes through the point $\big(\frac{4}{3}=0\big)$
$\therefore0-\text{y}_1\Big(\frac{3\text{x}_1}{\text{y}_1}\Big)\Big(\frac{4}{3}-\text{x}_1\Big)$
$\Rightarrow-\text{y}_1=\Big(\frac{3\text{x}_1}{\text{y}_1}\Big)\Big(\frac{4-3\text{x}_1}{3}\Big)$
$\Rightarrow-\text{y}_1^2=\text{x}_1(4-3\text{x}_1)$
$\Rightarrow-(3\text{x}_1^2-8)=\text{x}_1(4-3\text{x}_1)$...(From equation (1))
$\Rightarrow3\text{x}_1^2+8=4\text{x}_1^2-3\text{x}_1^2$
$\Rightarrow4\text{x}_1=8$
$\Rightarrow\text{x}_1=\frac{8}{4}$
$\Rightarrow\text{x}_1=2$
Substituting the value $\text{x}_1=2$ in equation (1), we have,
$\text{y}_1{^2}=3\times2^2-8$
$\Rightarrow\text{y}_1{^2}=12-8$
$\Rightarrow\text{y}_1{^2}=4$
$\Rightarrow\text{y}_1=\pm2$
Equation of the tangent is
$\text{y}-\text{y}_1=\frac{3\text{x}_1}{\text{y}_1}(\text{x}-\text{x}_1)$
Substituting the value of y1, we have,
$\text{y}-2=\frac{3\times2}{2}(\text{x}-2)$
and
$\text{y}+2=\frac{3\times2}{2}(\text{x}-2)$
Therefore, the equation are:
y - 2 = 3x - 6
and
y + 2 = 3x - 6
The point $\big(\frac{4}{3},0\big)$ does not satisfy the equation of the tangent,
y + 2 = 3x - 6
Thus, the equation of the tangent is y - 2 = 3x - 6
$\Rightarrow\text{y}=3\text{x}-4$
View full question & answer→Question 324 Marks
Find the equations of the tangent and the normal to the following curves at the indicated points.
$\text{xy}=\text{c}^2\text{ at }\big(\text{ct},\frac{\text{c}}{\text{t}}\big)$
Answer$\text{xy}=\text{c}^2$
Differentiating both sides w.r.t.x,
$\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-\text{y}}{\text{x}}$
Given $(\text{x}_1,\text{y}_1)=\Big(\text{ct},\frac{\text{c}}{\text{t}}\Big)$
Slope of tangent, $\text{m}=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\big(\text{ct},\frac{\text{c}}{\text{t}}\big)}=\frac{\frac{\text{c}}{\text{t}}}{\text{ct}}=\frac{-1}{\text{t}^2}$
Equation of tangent is,
$\text{y}-\text{y}_1=\text{m}(\text{x}-\text{x}_1)$
$\Rightarrow\text{y}-\frac{\text{c}}{\text{t}}=\frac{-1}{\text{t}^2}(\text{x}-\text{ct})$
$\Rightarrow\frac{\text{yt}-\text{c}}{\text{t}}=\frac{-1}{\text{t}^2}(\text{x}-\text{ct})$
$\Rightarrow\text{yt}^2-\text{ct}=-\text{x}+\text{ct}$
$\Rightarrow\text{x}+\text{yt}^2=2\text{ct}$
Equation of normal is,
$\text{y}-\text{y}_1=\frac{-1}{\text{m}}(\text{x}-\text{x}_1)$
$\Rightarrow\text{y}-\frac{\text{c}}{\text{t}}=\text{t}^2(\text{x}-\text{ct})$
$\Rightarrow\text{yt}-\text{c}=\text{t}^3\text{x}-\text{ct}^4$
$\Rightarrow\text{xt}^3-\text{yt}=\text{ct}^4-\text{c}$
View full question & answer→Question 334 Marks
Find the equation of the tangent line to the curve y = x2 - 2x + 7 which is parallel to the line 2x - y + 9 = 0
AnswerSlope of the given line is 2
Let $(\text{x}_1,\text{y}_1)$ be the point where the tangent is drawn to the curve $\text{y}=\text{x}^2-2\text{x}+7$
Since, the point lie on the curve.
Hence, $\text{y}=\text{x}_1^2-2\text{x}+7$
Now, $\text{y}=\text{x}^2-2\text{x}+7$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=2\text{x}-2$
Slope of the tangent at point $(\text{x}_1,\text{y}_1)$ = 2x - 2
Given that
Slope of the tangent = Slope of the given line
$\Rightarrow2\text{x}_1-2=2$
$\Rightarrow2\text{x}_1=4$
$\Rightarrow\text{x}_1=2$
Now, $\text{y}_1=4-4+7=7$
$\therefore(\text{x}_1,\text{y}_1)=(2,7)$
Equation of tangent is,
$\text{y}-\text{y}_1=\text{m}(\text{x}-\text{x}_1)$
$\Rightarrow\text{y}-7=2(\text{x}-2)$
$\Rightarrow\text{y}-7=2\text{x}-2$
$\Rightarrow2\text{x}-\text{y}+3=0$
View full question & answer→Question 344 Marks
Find the points on the curve $\frac{\text{x}^2}{4}+\frac{\text{y}^2}{25}=1$ at which the tangent are parallel to the
- x-axis
- y-axis.
AnswerDifferentiating $\frac{\text{x}^2}{4}+\frac{\text{y}^2}{25}=1$ with respect to x, we get $\frac{\text{x}}{2}+\frac{2\text{y}}{25}.\frac{\text{dy}}{\text{dx}}=0$ or $\frac{\text{dy}}{\text{dx}}=\frac{-25}{4}.\frac{\text{x}}{\text{y}}$ - Now, the tangent is parallel to the x-axis if the slope of the tangent is zero.
$\therefore\frac{-25}{4}.\frac{\text{x}}{\text{y}}=0$
This is possible if x = 0.
Then $\frac{\text{x}^2}{4}+\frac{\text{y}^2}{25}=1$ for x = 0 gives $\text{y}^2=25$
$\therefore\text{y}=\pm5$
Thus, the point at which the tangent are parallel to the x-axis are (0, 5) and (0, -5).
- Now, the tangent is parallel to the y-axis if the slope of the normal is zero.
$\therefore\frac{4\text{y}}{25\text{x}}=0$
This is possible is y = 0.
Then $\frac{\text{x}^2}{4}+\frac{\text{y}^2}{25}=1$ for y = 0 gives $\text{x}^2=4$
$\therefore\text{X}=\pm2$
Thus, the points at which the tangents are parallel to the y-axis are (2, 0) and (-2, 0)
View full question & answer→Question 354 Marks
Find the equations of the tangent and the normal to the following curves at the indicated points.
y = x2 at (0, 0)
AnswerThe equation of the curve is y = $\text{x}^2$
On differentiating with respect to x, we get:
$\frac{\text{dy}}{\text{dx}}=2\text{x}$
$\frac{\text{dy}}{\text{dx}}\Big]_{(0,\ 0)}=0$
Thus, the slope of the tangent at (0, 0) is 0 and the equation of the tangent is given as:
$\text{y}-0=0(\text{x}-0)$
$\Rightarrow\text{y}=0$
The slope of the normal at (0, 0) is $\frac{-1}{\text{slope of the tangent at (0,0)}}=-\frac{1}{0},$ which is not defined.
Therefore, the equation of the normal at $(\text{x}_0,\text{y}_0)=(0,0)$ is given by $\text{x}=\text{x}_0=0.$
View full question & answer→Question 364 Marks
Find the angle of intersecting of the following curves:
$\text{y}=\text{x}^2\text{ and }\text{x}^2+\text{y}^2=20$
Answerwe know that angle of intersecting of two curves is given by
$\tan\theta=\Big|\frac{\text{m}_1-\text{m}_2}{1+\text{m}_1\text{m}_2}\Big|...(\text{A})$
Where m1 and m2 are slopes of curves.
y = x2 ...(i)
x2 + y2 = 20 ...(ii)
Solving (i) and (ii)
y + y2 = 20
⇒ y + y2 - 20 = 0
⇒ (y + 5) (y - 4) = 0
⇒ y = -5, 4
$\therefore\text{x}=\sqrt{-5},\pm2$
$\therefore$ point are p = (2, 4), q = (-2, 4)
Now,
slope m1 for (i)
m1 = 2x = 4
slope m2 for (ii)
$\text{m}_2=\frac{\text{dy}}{\text{dx}}=\frac{-\text{x}}{\text{y}}=\frac{-1}{2}$
Now,
$\tan\theta=\Big|\frac{\text{m}_2-\text{m}_1}{1+\text{m}_1\text{m}_2}\Big|=\Bigg|\frac{\frac{-1}{2}-4}{1-\frac{1}{2}\times4}\Bigg|=\frac{9}{2}$
$\therefore\theta=\tan^{-1}\frac{9}{2}$
View full question & answer→Question 374 Marks
Find the condition for the following set of curves to intersect orthogonally
$\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}=1\text{ and }\frac{\text{x}^2}{\text{A}^2}-\frac{\text{y}^2}{\text{B}^2}=1$
AnswerWe have,
$\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}=1\ ...(1)$
$\frac{\text{x}^2}{\text{A}^2}-\frac{\text{y}^2}{\text{B}^2}=1\ ...(2)$
Slope if (1)
$\frac{2\text{x}}{\text{a}^2}+\frac{2\text{y}}{\text{b}^2}\times\frac{\text{dy}}{\text{dx}}=0$
$\therefore\text{m}_1=\frac{\text{dy}}{\text{dx}}=-\frac{\text{x}}{\text{y}}\frac{\text{b}^2}{\text{a}^2}$
Slope if (2)
$\frac{2\text{x}}{\text{A}^2}-\frac{2\text{y}}{\text{B}^2}\times\frac{\text{dy}}{\text{dx}}=0$
$\therefore\text{m}_2=\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\text{y}}\times\frac{\text{B}^2}{\text{A}^2}=-1$
(1) and (2) cuts orthogonally
$\therefore\text{m}_1\times\text{m}_2=-1$
$\therefore\frac{-\text{x}}{\text{y}}\frac{\text{b}^2}{\text{a}^2}\times\frac{\text{x}}{\text{y}}\times\frac{\text{B}^2}{\text{A}^2}=-1$
$\Rightarrow\frac{\text{x}^2}{\text{y}^2}\times\frac{\text{b}\text{B}^2}{\text{a}^2\text{A}^2}=1$
$\Rightarrow\frac{\text{x}^2}{\text{y}^2}\times\frac{\text{a}^2\text{A}^2}{\text{b}^2\text{B}^2}\ ...(3)$
Now,
(1) - (2) gives
$\text{x}^2\big[\frac{1}{\text{a}^2}-\frac{1}{\text{A}^2}\big]+\text{y}^2\big[\frac{1}{\text{b}^2}+\frac{1}{\text{B}^2}\Big]=0$
Put in (3) we get,
$\frac{(\text{B}^2+\text{b}^2)}{\text{b}^2\text{B}^2}\times\frac{\text{a}^2\text{A}^2}{(\text{a}^2-\text{A}^2 )}=\frac{\text{a}^2\text{A}^2}{\text{b}^2\text{B}^2}$
$\Rightarrow\text{B}^2+\text{b}^2=\text{a}^2-\text{A}^2$
$\Rightarrow\text{a}^2+\text{b}^2=\text{A}^2-\text{B}^2$
View full question & answer→Question 384 Marks
Find the slopes of the tangent and the normal to the following curves at the indicated points:
$\text{y}=(\sin2\text{x}+\cot\text{x}+2)^2\text{at}\text{ x}=\frac{\pi}{2}$
AnswerWe know that the slope of the tangent to the curve y = f(x) is
$\frac{\text{dy}}{\text{dx}}=\text{f}'(\text{x})\ ...(\text{A})$
And the slope of the normal is
$\frac{-1}{\frac{\text{dy}}{\text{dx}}}=\frac{-1}{\text{f}'(\text{x})}\ ...(\text{B})$
$\text{y}=(\sin2\text{x}+\cot\text{x}+2)^2$
$\therefore\frac{\text{dy}}{\text{dx}}=2(\sin2\text{x}+\cot\text{x}+2)(2\cos2\text{x}-\text{cosec}^2\text{x})$
$\therefore$ slope of tangent of $\text{x}=\frac{\pi}{2}$ is
$\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=\frac{\pi}{2}}=2\big(\sin\pi+\cos\frac{\pi}{2}+2\big)\big(2\cos\pi-\text{cosec}^2\frac{\pi}{2}\big)$
$=2(0+0+2)(-2-1)$
$=-12$
$\therefore$ slope of the normal is
$\frac{-1}{\frac{\text{dy}}{\text{dx}}}=\frac{1}{12}$
View full question & answer→Question 394 Marks
Find the points on the curve xy + 4 = 0 at which the tangents are inclined at an angle of 45° with the x-axis.
Answerxy + 4 = 0 ...(1)
Since the point satisfi the above equation,
x1y1 + 4 = 0 ...(2)
On differentiating equation (2) both sides with respwct to x, we get
$\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-\text{y}}{\text{x}}$
Slope of the tangent at $(\text{x}_1,\text{ y}_1)=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(\text{x},\text{ y})}=\frac{-\text{y}_1}{\text{x}_1}$
Slope of the tangent = 1 [given]
$\therefore\frac{-\text{y}_1}{\text{x}_1}=1$
View full question & answer→Question 404 Marks
Find the slopes of the tangent and the normal to the following curves at the indicated points:
$\text{x}=\text{a}(\theta-\sin\theta),\text{y}=\text{a}(1-\cos\theta)\text{at}\theta=-\frac{\pi}{2}$
Answer$\text{x}=\text{a}\cos^3\theta$
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=-3\text{a}\cos^2\theta\sin\theta$
$\text{y}=\text{a}\sin^3\theta$
$\Rightarrow\frac{\text{dy}}{\text{d}\theta}=3\text{a}\sin^2\cos\theta$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}=\frac{3\text{a}\sin^2\theta\cos\theta}{-3\text{a}\cos^2\theta\sin\theta}=-\tan\theta$
Now,
slope of the tangent $=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\theta=\frac{\pi}{4}}=-\tan\frac{\pi}{4}=-1$
slope of the normal $=\frac{-1}{\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\theta=\frac{\pi}{4}}}=\frac{-1}{-1}=1$
View full question & answer→Question 414 Marks
Find the points on the curve 2a2y = x3 - 3ax2 where the tangent is parallel to x-axis.
AnswerLet (x1, y1) represent the required points.
The slope of the x-axis is 0.
Here,
$2\text{a}^2\text{y}=\text{x}^3-\text{ax}^2$
Since, the point lies on the curve.
Hence, $2\text{a}^2\text{y}_1=\text{x}_1^3-3\text{ax}_1^2\ ...(1)$
On differentiating bith sides w.r.t. x, we get
$2\text{a}^2\frac{\text{dy}}{\text{d}}=3\text{x}^2-6\text{ax}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{3\text{x}^2-6\text{ax}}{2\text{a}^2}$
Slope of the tangent at $(\text{x}_1,\text{y}_1)=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(\text{x}_1,\text{ y}_1)}=\frac{3\text{x}_1^2-6\text{ax}_1}{2\text{a}^2}$
Given:
Slope of the tangent at (x1, y1) = slope of the x axis
$\Rightarrow\frac{3\text{x}_1{^2-6\text{ax}_1}}{2\text{a}}=0$
$\Rightarrow3\text{x}_1^2-6\text{ax}_1=0$
$\Rightarrow\text{x}_1(3\text{x}_1-6\text{a})=0$
$\Rightarrow\text{x}_1=0\text{ or }\text{x}_1=2\text{a}$
Also,
$2\text{a}^2\text{y}_1=0\text{ or }2\text{a}^2\text{y}_1=8\text{a}^3-12\text{a}^3$ [From eq. (1)]
$\Rightarrow\text{y}_1=0\text{ or }\text{y}_1=-2\text{a}$
Thus, the required point are (0, 0) and (2a, -2a).
View full question & answer→Question 424 Marks
Find the angle of intersecting of the following curves:
$\text{x}^2+\text{y}^2=2\text{x}\text{ and }\text{y}^2=8\text{x}$
AnswerWe know that angle of intersection of two curves is given by
$\tan\theta=\Big|\frac{\text{m}_1-\text{m}_2}{1+\text{m}_1\text{m}_2}\Big|...(\text{A})$
Where m1 and m2 are slope of curve
x2 + y2 = 2x ...(i)
y2 = x ... (ii)
Solving (i) and (ii)
x2 + x = 2x
⇒ x2 - x = 0
⇒ x (x - 1) = 0
⇒ x = 0, 1
$\therefore$ y = 0 or 1
$\therefore$ The point of intersection is P = (0, 0), Q = (1, 1)
$\therefore$ slope of (i)
$2\text{y}\frac{\text{dy}}{\text{dx}}=2-2\text{x}$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{2-2\text{x}}{2\text{y}}=\frac{1-\text{x}}{\text{y}}$
$\therefore\text{m}_1=0$
$\therefore$ slope of (ii)
$\text{m}_2=\frac{1}{2\text{y}}=\frac{1}{2}$
From (A)
$\tan\theta=\Big|\frac{\frac{1}{2}-0}{1+\frac{1}{2}\times0}\Big|=\frac{1}{2}$
$\therefore\theta=\tan^{-1}\Big(\frac{1}{2}\Big)$
View full question & answer→Question 434 Marks
Find the equations of the tangent and the normal to the following curves at the indicated points.
$\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}=1\text{ at }(\text{x}_1,\text{y}_1)$
Answerwe know that the equation of tangent and the normal to any curve is given by
$\text{y}-\text{y}_1=\text{m}(\text{x}-\text{x}_1)\ (1)\ \text{Tangent}$
$\text{y}-\text{y}_1=\frac{-1}{\text{m}}(\text{x}-\text{x}_1)\ (2)\ \text{Normal}$
Where m is the slope
$\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1\ ...(1)$
Differentiating with resect to x, we get
$\frac{2\text{x}}{\text{a}^2}+\frac{2\text{y}}{\text{b}^2}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{xb}^2}{\text{ya}^2}$
$\therefore\text{slope}\text{ m }=\Big(\frac{\text{dy}}{\text{dx}}\Big)_\text{p}=-\frac{\text{x}_1\text{b}^2}{\text{y}_1\text{a}^2}$
From (1)
Equation of tangent is
$(\text{y}-\text{y}_1)=-\frac{\text{x}_1\text{b}^2}{\text{y}_1\text{a}^2}(\text{x}-\text{x}_1)$
$\Rightarrow\text{xx}_1\text{b}^2+\text{yy}_1\text{a}^2=\text{x}_1^2\text{b}^2+\text{y}_1^2\text{a}^2$
Dividing by a2b2 both side
$\Rightarrow\frac{\text{xx}_1}{\text{a}^2}+\frac{\text{yy}_1}{\text{b}^2}=\frac{\text{x}_1{^2}}{\text{a}^2}+\frac{\text{y}_1{^2}}{\text{b}^2}=1\ [\because(\text{x}_1,\text{y}_1)\text{lies on}(1)]$
$\therefore\frac{\text{xx}_1}{\text{a}^2}+\frac{\text{yy}_1}{\text{b}^2}=1$
From (2)
Equation of normal is
$(\text{y}-\text{y}_1)=\frac{\text{y}_1\text{a}^2}{\text{x}_1\text{b}^2}(\text{x}-\text{x}_1)$
$\text{xy}_1\text{a}^2-\text{yx}_1\text{b}^2=\text{x}_1\text{y}_1\text{a}^2-\text{y}_1\text{x}_1\text{b}^2$
Dividing by x1y1 both side
$\frac{\text{xa}^2}{\text{x}_1}-\frac{\text{yb}^2}{\text{y}_1}=\text{a}^2-\text{b}^2$
View full question & answer→Question 444 Marks
Find the equations of the tangent and the normal to the following curves at the indicated points.
$\text{y}^2=\frac{\text{x}^3}{4-\text{x}}\text{ at }(2,-2)$
Answer$\text{y}^2=\frac{\text{x}^3}{4-\text{x}}$
Differentiating both sides w.r.t.x,
$2\text{y}\frac{\text{dy}}{\text{dx}}=\frac{(4-\text{x})(3\text{x}^2)-\text{x}^3(-1)}{(4-\text{x})^2}=\frac{12\text{x}^2-3\text{x}^3+\text{x}^3}{(4-\text{x})^2}=\frac{12\text{x}^2-2\text{x}^3}{(4-\text{x})^2}$
$\frac{\text{dy}}{\text{dx}}=\frac{12\text{x}^2-2\text{x}^3}{2\text{y}(4-\text{x})^2}$
Given $(\text{x}_1,\text{y}_1)=(2,-2)$
Slope of tangent, $\text{m}=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(2,-2)}=\frac{48-16}{-16}=-2$
Equation of tangent is,
$\text{y}-\text{y}_1=\text{m}(\text{x}-\text{x}_1)$
$\Rightarrow\text{y}+2=-2(\text{x}-2)$
$\Rightarrow\text{y}+2=-2\text{x}+4$
$\Rightarrow2\text{x}+\text{y}-2=0$
Equation of normal is, $\text{y}-\text{y}_1=\frac{-1}{\text{m}}(\text{x}-\text{x}_1)$
$\Rightarrow\text{y}+2=\frac{1}{2}(\text{x}-2)$
$\Rightarrow2\text{y}+4=\text{x}-2$
$\Rightarrow\text{x}-2\text{y}-6=0$
View full question & answer→Question 454 Marks
Find the points on the curve y = x3 where the slope of the tangent is equal to the x-coordinate of the point.
AnswerLet (x1, y1) be the required point.
x coordinate of the point is x1.
Since, the point lies on the curve.
Hence, $\text{y}_1=\text{x}_1^3\ ...(1)$
Now, $\text{y}=\text{x}^3$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=3\text{x}^2$
Slope of tangent at (x, y) $=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(\text{x}_1,\text{ y}_1)}=3\text{x}_1^2$
Given that
Slope of the tangent is equal to the x-coordinate of the point.
$\Rightarrow3\text{x}_1^2=\text{x}_1$
$\Rightarrow\text{x}_1(3\text{x}_1-1)=0$
$\Rightarrow\text{x}_1=0\text{ or }\text{x}_1=\frac{1}{3}$
$\Rightarrow\text{y}_1=0^3\text{ or }\text{y}_1=\Big(\frac{1}{3}\Big)^3$[from (1)]
$\Rightarrow\text{y}_1=0\text{ or }\text{y}_1=\frac{1}{27}$
So, the point are $(\text{x}_1,\text{y}_1)=(0,0),\Big(\frac{1}{3},\frac{1}{27}\Big)$
View full question & answer→Question 464 Marks
Find the equations of the tangent and the normal to the following curves at the indicated points.
$\text{y}^2=4\text{x}\text{ at }(1,2)$
Answer$\text{y}^2=4\text{x}$
Differentiating both sides w.r.t.x,
$\text{2}\text{y}\frac{\text{dy}}{\text{dx}}=4$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2}{\text{y}}$
Slope of the tangent, $\text{m}=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(1,2)}=\frac{2}{2}=1$
Given $(\text{x}_1,\text{y}_1)=(1,2)$
Equation of tangent is,
$\text{y}-\text{y}_1=\text{m}(\text{x}-\text{x}_1)$
$\Rightarrow\text{y}-2=1(\text{x}-1)$
$\Rightarrow\text{y}-2=\text{x}-1$
$\Rightarrow\text{x}+\text{y}+1=0$
Equation of normal is,
$\text{y}-\text{y}_1=\frac{-1}{\text{m}}(\text{x}-\text{x}_1)$
$\Rightarrow\text{y}-2=1(\text{x}-1)$
$\Rightarrow\text{y}-2=-\text{x}+1$
$\Rightarrow\text{x}+\text{y}-3=0$
View full question & answer→Question 474 Marks
Find the slopes of the tangent and the normal to the following curves at the indicated points:
$\text{x}^2+3\text{y}+\text{y}^2=5\ \text{at}\ (1,1)$
Answer$\text{x}^2+3\text{y}+\text{y}^2=5$
On differentiating both sides w.r.t. x, we get
$2\text{x}+3\frac{\text{dy}}{\text{dx}}+2\text{y}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}(3+2\text{y})=-2\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-2\text{x}}{3+2\text{y}}$
Now,
Slope of the tangent $=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(1,1)}=\frac{-2\text{x}}{3+2\text{y}}=\frac{-2}{3+2}=\frac{-2}{5}$
Slope of the normal $=\frac{-1}{\big(\frac{\text{dy}}{\text{dx}}\big)_{(1,1)}}=\frac{-1}{\big(\frac{-2}{5}\big)}=\frac{5}{2}$
View full question & answer→Question 484 Marks
Find the equations of the tangent and the normal to the following curves at the indicated points.
$\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1\text{ at }(\text{a}\sec\theta,\text{b}\tan\theta)$
Answer$\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}=1$
Differentiating both sides w.r.t. x,
$\Rightarrow\frac{2\text{x}}{\text{a}^2}-\frac{2\text{y}}{\text{b}^2}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{2\text{y}}{\text{b}^2}\frac{\text{dy}}{\text{dx}}=\frac{2\text{x}}{\text{a}^2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{xb}^2}{\text{ya}^2}$
Slope of tangent, $\text{m}=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(\text{a}\sec\theta,\text{b}\tan\theta)}$
$=\frac{\text{a}\sec\theta(\text{b}^2)}{\text{b}\tan\theta(\text{a}^2)}=\frac{\text{b}}{\text{a}\sin\theta}$
Given:
$(\text{x}_1,\text{y}_1)=(\text{a}\sec\theta,\text{b}\tan\theta)$
Equation of tangent is,
$\text{y}-\text{y}_1=\text{m}(\text{x}-\text{x}_1)$
$\Rightarrow\text{y}-\text{b}\tan\theta=\frac{\text{b}}{\text{a}\sin\theta}(\text{x}-\text{a}\sec\theta)$
$\Rightarrow\text{ay}\sin\theta-\text{ab}\frac{\sin^2\theta}{\cos\theta}=\text{bx}-\frac{\text{ab}}{\cos\theta}$
Equation of tangent is,
$\text{y}-\text{y}_1=\text{m}(\text{x}-\text{x}_1)$
$\Rightarrow\text{y}-\text{b}\tan\theta=\frac{\text{b}}{\text{a}\sin\theta}(\text{x}-\text{a}\sec\theta)$
$\Rightarrow\text{ay}\sin\theta-\text{ab}\frac{\sin^2\theta}{\cos\theta}=\text{bx}-\frac{\text{ab}}{\cos\theta}$
$\Rightarrow\frac{\text{ay}\sin\theta\cos\theta-\text{ab}\sin^2\theta}{\cos\theta}=\frac{\text{bx}\cos\theta-\text{ab}}{\cos\theta}$
$\Rightarrow\text{ay}\sin\theta\cos\theta-\text{ab}\sin^2\theta=\text{bx}\cos\theta-\text{ab}$
$\Rightarrow\text{bx}\cos\theta-\text{ay}\sin\theta\cos\theta=\text{ab}(1-\sin^2\theta)$
$\Rightarrow\text{bx}\cos\theta-\text{ay}\sin\theta\cos\theta=\text{ab}\cos^2\theta$
Dividing by ab $\cos^2\theta,$
$\Rightarrow\frac{\text{x}}{\text{a}}\sec\theta-\frac{\text{y}}{\text{b}}\tan\theta=1$
Equation of normal is,
$\text{y}-\text{y}_1=\frac{-1}{\text{m}}(\text{x}-\text{x}_1)$
$\Rightarrow\text{y}-\text{b}\tan\theta=\frac{-\text{a}\sin\theta}{\text{b}}(\text{x}-\text{a}\sec\theta)$
$\Rightarrow\text{yb}-\text{b}^2\tan\theta=-\text{ax}\sin\theta+\text{a}^2\tan\theta$
$\Rightarrow\text{ax}\sin\theta+\text{by}=(\text{a}^2+\text{b}^2)\tan\theta$
Dividing by $\tan\theta$
$\text{ax}\cos\theta+\text{by}\cot\theta=(\text{a}^2+\text{b}^2)$
View full question & answer→Question 494 Marks
Find the angle of intersecting of the following curves:
$\text{y}=4-\text{x}^2\text{ and }\text{y}=\text{x}^2$
AnswerGiven curve are, y = 4 - x2 ...(1) y = x2 ...(2) From (1) and (2), we get $4-\text{x}^2=\text{x}^2$ $\Rightarrow2\text{x}^2=4$ $\Rightarrow\text{x}^2=2$ $\Rightarrow\text{x}=\pm\sqrt{2}$ Substituting the value of x in (2), we get, ⇒ y = 2 $\Rightarrow(\text{x},\text{y})=(\sqrt{2},2),(-\sqrt{2},2)$ Differentiating (1) w.r.t.x, $\frac{\text{dy}}{\text{dx}}=-2\text{x}\ ...(3)$ Differentiating (2) w.r.t.x, $\frac{\text{dy}}{\text{dx}}=-2\text{x}\ ...(4)$ case 1: $(\text{x},\text{y})=(\sqrt{2},2)$ From (3), we have, $\text{m}_1=-2\sqrt{2}$ From (4), we have, $\text{m}_2=2\sqrt{2}$ Now, $\tan\theta=\Big|\frac{\text{m}_1-\text{m}_2}{1+\text{m}_1\text{m}_2}\Big|=\Big|\frac{-2\sqrt{2}-2\sqrt{2}}{1-8}\Big|=\frac{4\sqrt{2}}{7}$ $\Rightarrow\theta=\tan^{-1}\Big(\frac{4\sqrt{2}}{7}\Big)$ Case 2:
$(\text{x},\text{y})=(-\sqrt{2},2)$ From (3), we have, $\text{m}_1=2\sqrt{2}$ From (4), we have, $\text{m}_2=-2\sqrt{2}$ Now, $\tan\theta=\Big|\frac{\text{m}_1-\text{m}_2}{1+\text{m}_1\text{m}_2}\Big|=\Big|\frac{-2\sqrt{2}+2\sqrt{2}}{1-8}\Big|=\frac{4\sqrt{2}}{7}$ $\Rightarrow\theta=\tan^{-1}\Big(\frac{4\sqrt{2}}{7}\Big)$ View full question & answer→Question 504 Marks
The equation of the tangent at (2, 3) on the curve y2 = ax3 + b is y = 4x − 5. Find the values of a and b.
AnswerThe slope of the given line y = 4x - 5 is 4
$\text{y}^2=\text{ax}^3+\text{b}...(1)$
$2\text{y}\frac{\text{dy}}{\text{dx}}=3\text{ax}^2$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{3\text{ax}^2}{2\text{y}}$
Slope of tangent $=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(2,3)}=\frac{12\text{a}}{6}=2\text{a}$
Given that
Slope of tangent = slope of given line
2a = 4
⇒ a = 2
Substituting this and x = 2, y = 3 in (1), we get
9 = 16 + b
⇒ b = -7
Hence, a = 2 and b = -7
View full question & answer→