Question 15 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 25 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.
Answerwe 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 35 Marks
Find a point on the curve $y = x^3 - 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 45 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 $(x_1, y_1)$ 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 55 Marks
Find the equation of the tangent line to the curve $y = x^2 - 2x + 7$ which is perpendicular to the line $5y - 15x = 13.$
AnswerSlope of the given line is 3
Slope of the line perpendicualr to this line $=\frac{-1}{3}$
Let $(\text{x}_1,\text{ y}_1)$ be the point where the tangent is drawn to the curve.
Since, the point lie on the curve.
Hence, $\text{y}=\text{x}_1{^2}-2\text{x}+7\ ...(1)$
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 $(\text{x}_1,\text{y}_1)$ = Slope of the perpendicular line.
$\Rightarrow2\text{x}_1-2=\frac{-1}{3}$
$\Rightarrow6\text{x}_1-6=-1$
$\Rightarrow6\text{x}_1=5$
$\Rightarrow\text{x}_1=\frac{5}{6}$
Now, $\text{y}_1=\frac{25}{36}-\frac{10}{6}+7=\frac{25-60+252}{36}=\frac{217}{36}$
$\therefore(\text{x}_1,\text{y}_1)=\Big(\frac{5}{6},\frac{217}{36}\Big)$
equation of tangent is,
$\text{y}-\frac{217}{36}=\frac{-1}{3}(\text{x}-\frac{5}{6})$
$\Rightarrow\frac{36\text{y}-217}{36}=\frac{-6\text{x}+5}{18}$
$\Rightarrow36\text{y}-217=-12\text{x}+10$
$\Rightarrow12\text{x}+36\text{y}-227=0$
View full question & answer→Question 65 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 75 Marks
Find the points on the curve $y = 3x^2 - 9x + 8$ at which the tangents are equally inclined with the axes.
AnswerThe given equation of the curve is
$\text{y}=3\text{x}^2-9\text{x}+8\ ...(1)$
Slope $=\frac{\text{dy}}{\text{dx}}=6\text{x}-9\ ...(2)$
As per question
The tangent is equally inclined to the axis
$\therefore\theta=\frac{\pi}{4}\text{ or }\frac{-\pi}{4}$
$\therefore$ Slope $=\tan\theta$
$=\tan\frac{\pi}{4}\ \text{or}\ \tan\Big(\frac{-\pi}{4}\Big)$
$=1\ \text{or}-1...(3)$
From (2) and (3), we have,
$6x - 9 = 1$ or $6x - 9 = -1$
$\Rightarrow\text{x}=\frac{5}{3}\ \text{or}\ \text{y}=\frac{4}{3}$
Thus, the points are
$\Big(\frac{5}{3},\frac{4}{3}\Big)\ \text{or}\Big(\frac{4}{3},\frac{4}{3}\Big)$
View full question & answer→Question 85 Marks
Find the equations of the tangent and the normal to the following curves at the indicated points.
$\text{x}=\text{at}^2,\text{y}=2\text{at at}\text{ t}=1$
Answer$\text{x}=\text{at}^2$ and $\text{y}=2\text{at}$
$\frac{\text{dx}}{\text{dt}}=2\text{at}$ and $\frac{\text{dy}}{\text{dt}}=2\text{a}$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{2\text{a}}{2\text{at}}=\frac{1}{\text{t}}$
Slope of tangent, $\text{m}=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{t}=1}=\frac{1}{1}=1$
$(\text{x}_1,\text{y}_1)=(\text{a},2\text{a})$
Equation of tangent is,
$\text{y}-\text{y}_1=\text{m}(\text{x}-\text{x}_1)$
$\Rightarrow\text{y}-2\text{a}=1(\text{x}-\text{a})$
$\Rightarrow\text{y}-2\text{a}=\text{x}-\text{a}$
$\Rightarrow\text{x}-\text{y}+\text{a}=0$
Equation of normal is, $\text{y}-\text{y}_1=\text{m}(\text{x}-\text{x}_1)$
$\Rightarrow\text{y}-2\text{a}=-1(\text{x}-\text{a})$
$\Rightarrow\text{y}-2\text{a}=-\text{x}+\text{a}$
$\Rightarrow\text{x}+\text{y}=3\text{a}$
View full question & answer→Question 95 Marks
Find the equations of the tangent and the normal to the following curves at the indicated points.
$\text{x}^2=4\text{y}\text{ at }(2,1)$
Answerwe know taht 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
We have,
$\text{x}^2=4\text{y}$
$\therefore2\text{x}=\frac{4\text{dy}}{\text{dx}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{2}$
$\therefore$ slope $\text{m}=\Big(\frac{\text{dy}}{\text{dx}}\Big)=1$
From (A)
Equation of tangent is
y - 1 = 1 (x - 2)
⇒ x - y = 1
From (B)
Equation of normal is
(y - 1) = -1 (x - 2)
⇒ x + y = 3
View full question & answer→Question 105 Marks
Find the equations of the tangent and the normal to the following curves at the indicated points.
$y = x^4 - bx^3 + 13x^2 - 10x + 5$ at $(0, 5)$
Answer$y = x^4 - bx^3 + 13x^2 - 10x + 5$ at $(0, 5)$
Differentiating both sides w.r.t.x,
$\frac{{\text{dy}}}{\text{dx}}=4\text{x}^3-3\text{bx}^2+26\text{x}-10$
Slope of tangent $,\text{m}=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(0,5)}=-10$
Given $(\text{x}_1,\text{y}_1)=(0,5)$
Equation of tangent is,
$\text{y}-\text{y}_1=\text{m}(\text{x}-\text{x}_1)$
$\Rightarrow\text{y}-5=-10(\text{x}-0)$
$\Rightarrow\text{y}-5=-10\text{x}$
$\Rightarrow\text{y}+10\text{x}-5=0$
Equation of normal is,
$\text{y}-\text{y}_1=\frac{-1}{\text{m}}(\text{x}-\text{x}_1)$
$\Rightarrow\text{y}-5=\frac{1}{10}(\text{x}-0)$
$\Rightarrow10\text{y}-50=\text{x}$
$\Rightarrow\text{x}-10\text{y}+50=0$
View full question & answer→Question 115 Marks
Find the equations of the tangent and the normal to the following curves at the indicated points.
$\text{y}=\text{x}^2+4\text{x}+1\text{ at }\text{x}=3$
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
We have,
$\text{y}=\text{x}^2+4\text{x}+1$ and $\text{p}=(\text{x}=3)$
Slope $=\frac{\text{dy}}{\text{dx}}=2\text{x}+4$
$\therefore\text{m}=\Big(\frac{\text{dy}}{\text{dx}}\Big)_\text{p}=10$
From (1)
Equation of tangent is
$(\text{y}-22)=10(\text{x}-3)$
$\Rightarrow10\text{x}-\text{y}=8$
From (1)
Equation of normal is
$(\text{y}-22)=10(\text{x}-3)$
$\Rightarrow\text{x}+10\text{y}=223$
View full question & answer→Question 125 Marks
Find the equation of the tangent to the curve $\text{y}=\sqrt{3}\text{x}-2$ which is parallel to the 4x − 2y + 5 = 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}=\sqrt{3}\text{x}-2$
Since, the point lies on the curve.
Hence, $\text{y}_1=\sqrt{3\text{x}_1-2}\ ...(1)$
Now, $\text{y}=\sqrt{3\text{x}_1-2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{3}{2\sqrt{3\text{x}-2}}$
Slope of the tangent at $(\text{x}_1,\text{y}_1)=\frac{3}{2\sqrt{3\text{x}_1-2}}$
Given that
Slope of the tangent = slope of the give line
$\Rightarrow\frac{3}{2\sqrt{3\text{x}_1-2}}=2$
$\Rightarrow3=4\sqrt{3\text{x}_1-2}$
$\Rightarrow9=16{(3\text{x}_1-2})$
$\Rightarrow\frac{9}{16}=3\text{x}_1-2$
$\Rightarrow3\text{x}_1=\frac{9}{16}+2=\frac{9+32}{16}=\frac{41}{16}$
$\Rightarrow\text{x}_1=\frac{41}{48}$
Now, $\text{y}_1=\sqrt{\frac{123}{48}-2}=\sqrt{\frac{27}{48}}=\sqrt{\frac{9}{16}}=\frac{3}{4}$
$\therefore(\text{x}_1,\text{y}_1)=\Big(\frac{41}{48},\frac{3}{4}\Big)$
Equation of tangent is,
$\text{y}-\text{y}_1=\text{m}(\text{x}-\text{x}_1)$
$\Rightarrow\text{y}-\frac{3}4{}=2\Big(\text{x}-\frac{41}{48}\Big)$
$\Rightarrow\frac{4\text{y}-3}{4}=2\Big(\frac{48\text{x}-41}{48}\Big)$
$\Rightarrow24\text{y}-18=48\text{x}-41$
$\Rightarrow48\text{x}-24\text{y}-23=0$
View full question & answer→Question 135 Marks
Find the values of a and b if the slope of the tangent to the curve xy + ax + by = 2 at (1, 1) is 2.
AnswerDifferentiatinng with respect to x, we get
$\text{y}+\text{x}\frac{\text{dy}}{\text{dx}}+\text{a}+\text{b}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}(\text{x}+\text{b})=-(\text{a}+\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-(\text{a}+\text{y})}{\text{x}+\text{b}}$
$\therefore$ slope of the tangent $=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}-1,\text{y}-1}=\frac{-(\text{a}+1)}{\text{b}+1}=2$
$\Rightarrow-(\text{a}+1)=2\text{b}+2$
$\Rightarrow2\text{b}+\text{a}=-3...(1)$
Also, (1, 1) lies on the curve, so x = 1, y = 1 satisfies the equation
xy + ax + by = 2
⇒ 1+ a + b = 2
⇒ a + b = 1 ...(2)
solving (1) and (2), we get
a = 5, b = 4
View full question & answer→Question 145 Marks
Find the points on the curve $y^2 = 2x^3$^ at which the slope of the tangent is 3.
AnswerLet $(x_1, y_1)$ be the required point.
Given:
$\text{y}^2=2\text{x}^3$
Since $(x_1, y_1)$ lies on a curve, $\text{y}_1^2=2\text{x}_1^3\ ...(1)$
$\Rightarrow2\text{y}\frac{\text{dy}}{\text{dx}}=6\text{x}^2$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{6\text{x}^2}{2\text{y}}=\frac{3\text{x}^2}{\text{y}}$
Slope of the tangent at (x, y) $=\frac{3\text{x}_1^2}{\text{y}_1}$
Slope of the tangent = 3 [given]
$\therefore$ $\frac{3\text{x}_1^2}{\text{y}_1}=3\ ...(2)$
$\Rightarrow\text{y}_1=\text{x}_1^2$
On substituting the value of $y_1$ in eq. (1), we get
$\text{x}_1^4=2\text{x}_1^3$
$\Rightarrow\text{x}_1^3(\text{x}_1-2)=0$
$\Rightarrow\text{x}_1=0,2$
case 1
When $\text{x}_1=0,\text{y}_1=\text{x}^2=0.$
Thus, we get the point (0, 0).
But, it does not satisfy eq. (2).
So, we can ignore (0, 0).
Case 2
When $\text{x} _1=2,\text{y}_1=\text{x}_1^2=4.$ Thus, we get the point (2, 4).
View full question & answer→Question 155 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 165 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 175 Marks
Find the equations of all lines of slope zero and that are tangent to the curve $\text{y}=\frac{1}{\text{x}^2-2\text{x}+3}$
AnswerThe slope of the tangent to the given curve at any point (x, y) is given by,
$\frac{\text{dy}}{\text{dx}}=\frac{-(2\text{x}-1)}{(\text{x}^2-2\text{x}+3)^2}$
If the slope of the tangent is 0, then we have:
$\frac{-2(\text{x}-1)}{(\text{x}^2-2\text{x}+3)^2}=0$
$\Rightarrow-2(\text{x}-1)=0$
$\Rightarrow\text{x}=1$
When x = 1, $\text{y}=\frac{1}{1-2+3}=\frac{1}{2}$
$\therefore$ The equation of the tangent through $\Big(1,\frac{1}{2}\Big)$ is given by,
$\text{y}-\frac{1}{2}=0(\text{x}-1)$
$\Rightarrow\text{y}-\frac{1}{2}=0$
$\Rightarrow\text{y}=\frac{1}{2}$
Hence, the equation of the required line is $\text{y}=\frac{1}{2}$
View full question & answer→Question 185 Marks
Find the equations of the tangent and the normal to the following curves at the indicated points.
$y = x^4 - 6x^3 + 13x^2 - 10x + 5$ at $x = 1$
Answer$y = x^4 − 6x^3 + 13x^2 − 10x + 5$ at $x = 1$
When $x = 1, y = 1 - 6 + 13 - 10 + 5 = 3$
So, $(x_1, y_1) = (1, 3)$
Now, $y = x^4 - 6x^3 + 13x^2 - 10x + 5$
Differentiating both sides w.r.t.x,
$\frac{\text{dy}}{\text{dx}}=4\text{x}^3-18\text{x}^3+26\text{x}-10$
Slope of the tangent, $\text{m}=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(1,\ 3)}=4-18+26-10=2$
Equation of tangent is,
$\text{y}-\text{y}_1=2(\text{x}-\text{x}_1)$
$\Rightarrow\text{y}-3=2(\text{x}-1)$
$\Rightarrow\text{y}-3=2\text{x}-2$
$\Rightarrow2\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}-3=\frac{-1}{2}(\text{x}-1)$
$\Rightarrow3\text{y}-6=-\text{x}+1$
$\Rightarrow\text{x}+3\text{y}-7=0$
View full question & answer→Question 195 Marks
Find the equations of the tangent and the normal to the following curves at the indicated points.
$\text{y}^2=4\text{ax}\text{ at }\Big(\frac{\text{a}}{\text{m}^2},\frac{2\text{a}}{\text{m}}\Big)$
Answerwe know taht 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{y}-\text{y}_1=\frac{-1}{\text{m}}(\text{x}-\text{x}_1)\ ...(2)$
Where m is the slope
We have
$\text{y}^2=4\text{ ax}\text{p}\Big(\frac{\text{a}}{\text{m}^2},\frac{2\text{a}}{\text{m}}\Big)$
Differentiating with respect to x, we get
$2\text{y}\frac{\text{dy}}{\text{dx}}=4\text{a}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2\text{a}}{\text{y}}$
$\therefore$ Slope $\text{m}=\Big(\frac{\text{dy}}{\text{dx}}\Big)_\text{p}=\text{m}$
From (1)
Equation of tangent is
$\Big(\text{y}-\frac{2\text{a}}{\text{m}}\Big)=\text{m}\Big(\text{x}-\frac{\text{a}}{\text{m}^2}\Big)$
$\Rightarrow\text{m}^2\text{x}-\text{my}=2\text{a}-\text{a}$
$\Rightarrow\text{m}^2\text{x}-\text{my}=\text{a}$
From (2)
Equation of normal is
$\Big(\text{y}-\frac{2\text{a}}{\text{m}}\Big)=\frac{-1}{\text{m}}\Big(\text{x}-\frac{\text{a}}{\text{m}^2}\Big)$
$\Rightarrow(\text{my}-2\text{a})=\frac{-\text{m}^2\text{x}+\text{a}}{\text{m}^2}$
$\Rightarrow\text{m}^2\text{x}+\text{m}^3\text{y}=2\text{am}^2+\text{a}$
$\Rightarrow\text{m}^2\text{x}+\text{m}^3\text{y}-2\text{am}^2-\text{a}=0$
View full question & answer→Question 205 Marks
Find the points on the curve $x^2 + y^2 = 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 215 Marks
Show that the following curves intersect orthogonally at the indicated points:
$x^2 = 4y$ and $4y + x^2 = 8$ at $(2, 1)$
Answer$x^2 = 4y ...(1)$
$4y + x^2 = 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 225 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 235 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, $y_2 = x ...(1) x_2 = 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 245 Marks
At what points on the curve $y = 2x^2 - x + 1$ is the tangent parallel to the line $y = 3x + 4?$
AnswerLet $(x_1, y_1)$ 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 255 Marks
Show that the following curves intersect orthogonally at the indicated points:
$x^2 = y$ and $x^3 + 6y = 7$ at $(1, 1)$
Answerwe have,
$x^2 = y ...(1)$
$x^3 + 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 265 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 275 Marks
Show that the following set of curves intersect orthogonally.
$x^3 - 3xy^2 = -2$ and $3x^2y - y^3 = 2$
Answerwe know that two curves interesects orthogonally if
$\text{m}_1\times\text{m}_2=-1\ ...(\text{A})$
Where $m_1$ and $m_2$ 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 285 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 295 Marks
Find the points on the curve $x^2 + y^2 - 2x - 3 = 0$ at which the tangents are parallel to the x-axis.
AnswerLet $(x_1, y_1)$ be the required point.
Since the point lie on the curve.
Hence $\text{x}_1^2+\text{y}_1^2-2\text{x}_1-3=0\ ...(1)$
Now, $\text{x}^2+\text{y}^2-2\text{x}-3=0$
$\Rightarrow2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}-2=0$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{2-2\text{x}}{2\text{y}}=\frac{1-\text{x}}{\text{y}}$
Now,
Slope of the tangent $=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(\text{x}_1,\text{y}_1)}=\frac{1-\text{x}_1}{\text{Y}_1}$
Slope of the tangent = 0 (Given)
$\therefore\frac{1-\text{x}_1}{\text{y}_1}=0$
$\Rightarrow1-\text{x}_1=0$
$\Rightarrow\text{x}_1=1$
From (1), we get
$\text{x}_1^2+\text{y}_1^2-2\text{x}_1-3=0$
$\Rightarrow1+\text{y}_1^2-2-3=0$
$\Rightarrow\text{y}_1^2-4=0$
$\Rightarrow\text{y}_1=\pm2$
Hence, the point are (1, 2) and (1, -2)
View full question & answer→Question 305 Marks
Find the equation of the normal to the curve $x^2 + 2y^2 - 4x - 6y + 8 = 0$ at the point whose abscissa is 2.
AnswerThe given equation of curve is
$x^2 + 2y^2 - 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 + 2y^2 - 8 - 6y + 8 = 0$
$\Rightarrow 2y^2 - 6y + 4 = 0$
$\Rightarrow y^2 - 3y + 2 = 0$
$\Rightarrow (y - 2) (y - 1) = 0$
$\Rightarrow 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 315 Marks
Find the equation of a normal to the curve $y = x \log_e$ x which is parallel to the line $2x − 2y + 3 = 0.$
AnswerThe given equation are,
$\text{y}=\text{x}\log_\text{e}\text{x}\ ...(1)$
$2\text{x}-2\text{y}+3=0\ ...(2)$
Slope $m_1$ of (1)
$\text{m}_1=\frac{\text{dy}}{\text{dx}}=\log_\text{e}\text{x}+1$
Slope $m_2$ of (2)
$m_2 = 1$
slope of the normal to (1) is,
$-\frac{1}{\text{m}_1}=-\frac{1}{\log\text{x}+1}$
According to the question,
$-\frac{1}{\log\text{x}+1}=1$
$\log\text{x}=-2$
$\text{x}=\text {e}^{-2}$
From (1)
$\text{y}=\text{e}^{-2}\log_\text{e}\text{e}^{-2}$
Thus,
$\text{p}=(\text{e}^{-2}.\text{e}^{-2}\log_\text{e}\text{e}^{-2})$
So, the equation of normal is,
$\text{y}-\text{e}^{-2}\log_\text{e}\text{e}^{-2}=1(\text{x}-\text{e}^{-2})$
$\text{x}-\text{y}=3\text{e}^{-2}$
View full question & answer→Question 325 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 335 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
$x^2 + y^2 = 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),
$x^2+ (2x - 4)^2 - 4x - 1 = 0$
$\Rightarrow x^{2 +}4x^2+ 16 - 16x - 4x - 1 = 0$
$\Rightarrow 5x^2 - 20x + 15 = 0$
$\Rightarrow x^2 - x + 3 = 0$
$\Rightarrow (x - 3) (x - 1) = 0$
$\Rightarrow 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 345 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 355 Marks
Find the points on the curve $y = x^3 - 2x^2 - 2x$ at which the tangent lines are parallel to the line $y = 2x - 3$.
AnswerThe given equations are
$y = x^3 - 2x^2 - 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
$3x^2 - 4x - 2 = 2$
$\Rightarrow 3x^2 - 4x - 4 = 0$
$\Rightarrow 3x^2 - 6x - 2x - 4 = 0$
$\Rightarrow 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 365 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,
$2y^2 = x^3 ...(1)$
$y^2= 32x ...(2)$
From the two equations we get
$2(32x) = x^3$
$\Rightarrow 64x = x^3$
$\Rightarrow x(x^2 - 64) = 0$
$\Rightarrow x = 0, 8, -8$
Substituting the value of x in (2) we get,
$y_1 = 0, 16, -16$
$\therefore (x_1, y_1) = (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 375 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 385 Marks
Find the point on the curve $y = x^2 - 2x + 3$, where the tangent is parallel to x-axis.
AnswerThe slope of the x-axis is 0
Now, let $(x_1, y_1)$ 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 $(x_1, y_1) =$ slope of the x - axis
$= 2x_1 - 2 = 0$
$\Rightarrow x_1 = 1$
and
$y_1 = 1 - 2 + 3 = 2 [$From $(1)]$
$\therefore$ Required point $= (x_1, y_1) = (1, 2)$
View full question & answer→Question 395 Marks
If the tangent to the curve $y = x^3 + 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 = x^3 + 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 405 Marks
Show that the following set of curves intersect orthogonally.
$y = x^3$ and $6y = 7 - x^2$
Answer$y = x^3 ...(1)$
$6y = 7 - x^2 ...(2)$
From (1) and (2) we get,
$6x^3 = 7 - x^2$
$\Rightarrow 6x^3- x^2 - 7 = 0$
$x = 1$ satisfies this.
Dividing this by $x - 1,$ we get
$6x^2 + 7x + 7 = 0$
Discriminant =$ 7^2 - 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 415 Marks
Find the equation of the tangent line to the curve $y = x^2 + 4x - 16$ which is parallel to the line $3x - y + 1 = 0.$
AnswerThe given equation are,
$y = x^2 + 4x - 16$
$3x - y + 1 = 0$
Slope $m_1$_ if (i)
$\text{m}_1=\frac{\text{dy}}{\text{dx}}=2\text{x}$
Slope $m_2$ if (ii)
$m_2= 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 425 Marks
Prove that the curves $y^2 = 4x$ and $x^2 + y^2 - 6x + 1 = 0$ touch each other at the point $(1, 2).$
AnswerGiven:
Curves $y^2 = 4x ...(1)$
And $x^2 + y^2 - 6x + 1 = 0 ...(2)$
$\therefore$ The point of intersection of two curves is $(1, 2)$
First curve is $y^2 = 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 $x^2 + y^2 - 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, $m_1 = m_2 = 1$ at $(1, 2)$
So, given curve touch each other at $(1, 2)$
View full question & answer→Question 435 Marks
Find the equation of the tangents to the curve $3x^2 - y^2 = 8$, which passes through the point $\big(\frac{4}{3},0\big)$
AnswerConsider the equation of the curve
$3x^2 - y^2 = 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 $y_1$, 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 445 Marks
Find the equation of the tangent line to the curve $y = x^2 - 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 455 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 465 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 475 Marks
Find the equations of the tangent and the normal to the following curves at the indicated points.
$y = x^2$ 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 485 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 $m_1$ and $m_2$ are slopes of curves.
$y = x^2 ...(i)$
$x^2+ y^2 = 20 ...(ii)$
Solving (i) and (ii)
$y + y^2= 20$
$\Rightarrow y + y^2- 20 = 0$
$\Rightarrow (y + 5) (y - 4) = 0$
$\Rightarrow y = -5, 4$
$\therefore\text{x}=\sqrt{-5},\pm2$
$\therefore$ point are $p = (2, 4), q = (-2, 4)$
Now,
slope $m_1$ for (i)
$m_1= 2x = 4$
slope $m_2$ 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 495 Marks
Find the point on the curve $y = 3x^2 + 4 $ at which the tangent is perpendicular to the line whose slop is $-\frac{1}{6}$
AnswerThe given equation of curve is
$\text{y}=3\text{x}^2+4\ ...(1)$
Slope $=\text{m}_1=\frac{\text{dy}}{\text{dx}}=6\text{x}\ ...(2)$
Now,
The given slope $\text{m}_2=\frac{-1}{6}$
We have
Tangent to (1) is perpendicular to the tangent whose slope is $\frac{-1}{6}$
$\therefore\text{m}_1\times\text{m}_2=-1$
$\Rightarrow6\text{x}\times\frac{-1}{6}=-1$
$\Rightarrow\text{x}=1$
From (1)
$y = 7$
Thus, the required point is $(1, 7).$
View full question & answer→Question 505 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→