Maths M.C.Q (1 Marks)

2. $\Big(\frac{1}{4},\frac{1}{2}\Big)$

Solution:

Let the required point be $(\text{x}_1,\text{y}_2).$

The tangent makes an angle of 45° with the x-axis.

Slope of the tangent $=\tan 45^\circ=1$

Since, the point lies on the curve.

Hence, $\text{y}_{1}^2=\text{x}_1$

Now,

$\text{y}^2=\text{x}$

$\Rightarrow2\text{y}\frac{\text{dy}}{\text{dx}}=1$

$\Rightarrow\frac{\text{dy}}{\text{dx}}\approx\frac{1}{2\text{y}}$

Slope of the tangent $=\bigg(\frac{\text{dy}}{\text{dx}}\bigg)_{(\text{x}_1,\text{y}_1)}=\frac{1}{2\text{y}_1}$

Given:

$\frac{1}{2\text{y}_1}=1$

$\Rightarrow2\text{y}_1=1$

$\Rightarrow\text{y}_1=\frac{1}{2}$

Now,

$\text{x}_1=\text{y}^2_1-\big(\frac{1}{2}\big)^2-\frac{1}{4}$

$\therefore(\text{x}_1,\text{y}_1)=\Big(\frac{1}{4},\frac{1}{2}\Big)$

2. $\Big(\frac{1}{4},\frac{1}{2}\Big)$

Solution:

Let the required point be(x₁, y₁).

The tangent makes an angle of 45° with the x-axis,

$\therefore$ Slope of the tangen = tan 45° = 1

Since, the point lies on the curve.

Hence, $\text{y}^2=\text{x}_1$

Now, $\text{y}^2=\text{x}$

$\Rightarrow2\text{y}\frac{\text{dy}}{\text{dx}}=1$

$\frac{\text{dy}}{\text{dx}}=\frac{1}{2\text{y}}$

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,

$\text{x}_1=\text{y}_1^2=\Big(\frac{1}{2}\Big)^2=\frac{1}{4}$

$\therefore(\text{x}_1,\text{y}^2_1)=\Big(\frac{1}{4},\frac{1}{2}\Big)$

2. (1, 0)

Solution:

$\text{y}=\text{x}$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=1$

Let (x_2, y₂) be the required point.

Since, the point lies on the curve,

Hence, $\text{y}_1=\text{x}_1^2-3\text{x}_1+2$

Now, $\text{y}=\text{x}^2-3\text{x}+2$

$\therefore\frac{\text{dy}}{\text{dx}}=2\text{x}-3$

Slope of the perpendicular to this line.

$\therefore$ Slope of the tangent $=\frac{-1}{\text{slope of the line}}=\frac{-1}{1}=-1$

Now,

$2\text{x}_1-3=-1$

$\Rightarrow 2\text{x}_1=2$

$\Rightarrow\text{x}_1=2$

and

$\text{y}_1=\text{x}_1^2-3\text{x}_1+2=1-3+2=0 $

$\therefore(\text{x}_1,\text{y}_1)=(1,0)$

4. (1, 2), (1, 2)

Solution:

Let (x₁, y₁) be the required point.

Since, the point lie on the curve.

Hence, $\text{x}^2_1+\text{y}_1^2-2\text{x}_1-3=0 \ ...(1)$

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 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

$\text{x}_1^2+\text{y}_1^2-2\text{x}_1-3=0$

$\Rightarrow1+\text{y}_1^2-4=0$

$\Rightarrow\text{y}_1=\pm2$

So, the points are (1, 2) and (1, -2).

2. x − y = 0

Solution:

2y + x² = 3

$2\frac{\text{dy}}{\text{dx}}+2\text{x}=0$

$\frac{\text{dy}}{\text{dx}}=-\text{x}$

$\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(1,1)}=-1$

Slope of the normal = 1

Equation of the normal

y - 1 = x - 1

y = x

x - y = 0

3. Makes an acute angle with x-axis.

Solution:

We have, y = 2x⁷ + 3x + 5

$\frac{\text{dy}}{\text{dx}}=14\text{x}^6+3$

$\Rightarrow\frac{\text{dy}}{\text{dx}}>3$ ($\because$ x⁶ is always positive for any real value of x)

$\Rightarrow\frac{\text{dy}}{\text{dx}}>0$

So, $\tan\theta>0$

Hence, $\theta$ lies in first quadrant.

Thus, the tangent to the curve makes an acute angle with x-axis.

2. $\frac{6}{7}$

Solution:

$\text{x}=\text{t}^2+3\text{t}-$ and $\text{y}=2\text{t}^2-2\text{t}^2-2\text{t}-5$

$\frac{\text{dx}}{\text{dt}}=2\text{t}+3$ and $\frac{\text{dy}}{\text{dt}}=4\text{t}-2$

$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{4\text{t}-2}{2\text{t}+3}$

The given point is (2, -1).

$\therefore\text{x}=2$ and $\text{y=}-1$

Now,

$\text{t}^2+3\text{t}-8=2 $ and $2\text{t}^2-2\text{t}-5=-1$

Let u solve one of these to get the value of t.

$\text{t}^2+3\text{t}-10=0$ and $2\text{t}^2-2\text{t}-4=0$

$\Rightarrow(\text{t}+5)(\text{t}-2)=0$ and $(2\text{t}-2)(\text{t}-2)=0$

$\Rightarrow\text{t}=-5\text{ or }\text{t}=2$

These two have t = 2 as a comman solution.

$\therefore$ Slope of the tangent $= \Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{t}=2}=\frac{8-2}{4+3}=\frac{6}{7}$

3. x + 3y ± 8 = 0

Solution:

Since the normal is parallel to the given line, the equation of normal will be of the given form.

$\text{x}+3\text{y}=\text{k}$

$3\text{x}^2-\text{y}^2=8$

Let (x₁, y₁) be the point of intersection of the two curves.

Then,

$\text{x}_1+3\text{y}_1=\text{k} \ ...(1)$

$3\text{x}_1^2-\text{y}_1^2=8 \ ...(2)$

Now,$$ $3\text{x}^2-\text{y}^2=8$

On diffierentiating both sides w.r.t.x, we get

$6\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}=0$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{6\text{x}}{2\text{y}}=\frac{3\text{x}}{\text{y}}$

Slope of the normal, $=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(\text{x}_1,\text{y}_1)}=\frac{3\text{x}_1}{\text{y}_1}$

Slope of the normal, $\text{m}=\frac{-1}{\Big(\frac{3\text{x}}{\text{y}}\Big)}=\frac{-\text{y}_1}{3\text{x}_1}$

Given:

Slope of the normal = Slope of the given line

$\Rightarrow\frac{-\text{y}_1}{3\text{x}_1}=\frac{-1}{3}$

$\Rightarrow\text{y}_1=\text{x}_1\ ...(3)$

From (2), we get

$3\text{x}_1{^2}-\text{x}_1{^2}=8$

$\Rightarrow2\text{x}_1{^2}=8$

$\Rightarrow\text{x}_1{^2}=4$

$\Rightarrow\text{x}_1=\pm2$

Case 1

when $\text{x}_1=2$

From (3), we get

$\text{y}_1=\text{x}_1=2$

From (3), we get

2 + 3 (2) = k

⇒ 2 + 6 = k

⇒ k = 8

$\therefore$ Equation of the normal from (1)

⇒ x + 3y = 8

⇒ x + 3y - 8 = 0

⇒ -2 - 6 = k

⇒ k = -8

$\therefore$ Equation of the normal from (1)

⇒ x + 3y = -8

⇒ x + 3y - 8 = 0

From both the case, we get the equation of the normal as:

$\text{x}+3\text{y}\pm8=0$

1. $\Big(4,\frac{8}{3}\Big)$

Solution:

9y² = x³

$\Rightarrow18\text{y}\frac{\text{dy}}{\text{dx}}=3\text{x}^2$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}^2}{6\text{y}}=\text{slope of tangent}$

Slope of normal $=\frac{-6\text{y}}{\text{x}^2}$

Given that normal makes equal intercept on axes.

⇒ Slope of normal $=\pm1$

$\frac{-6\text{y}}{\text{x}^2}=1$ or $\frac{-6\text{y}}{\text{x}^2=}=-1$

$\Rightarrow\text{y}=\frac{-\text{x}^2}{6}$ or $\text{y}=\frac{\text{x}^2}{6}$

$\text{y}=\frac{-\text{x}^2}{6}$

$9\text{y}^2=\text{x}^3$

$9\Big(\frac{-\text{x}^2}{6}\Big)^2=\text{x}^3$

$\Rightarrow\text{x}=0\text{ or }4$

$\Rightarrow\text{y}=0,\pm\frac{8}{3}$

Point are $\Big(4,\frac{8}{3}\Big)$ and $\Big(4,\frac{-8}{3}\Big)$

3. (-1, 0), (1, 2)

Solution:

$\text{x}\text{y}=\text{a}^2$ and $\text{x}^2-\text{y}^2=2\text{a}^2$

$\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=0$ and $2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}=0$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-\text{y}}{\text{x}}$ and $\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\text{y}}$

We can see dearly that product of the slopes of tan gents is -1 

Hence, angle between two tangents is 90°.

3. 6

Solution:

Given:

$\text{ay}+\text{x}^2=7 \ ...(1)$

$\text{x}^3=\text{y} \ ...(2)$

Point = (1, 1)

On differentiatuing (1) w.r.t.x, we get

$\text{a}\frac{\text{dy}}{\text{dx}}+2\text{x}=0$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-2\text{x}}{\text{a}}$

$\Rightarrow\text{m}_1=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(1,1)}=\frac{-2}{\text{a}}$

Again, on differetiating (2) w.r.t.x, we get

$3\text{x}^2=\frac{\text{dy}}{\text{dx}}$

$\Rightarrow\text{m}_2=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(1,1)}=3$

It is given that curves are orthogonal at the given point,

$\therefore\text{m}_1\times\text{m}_2=-1$

$\Rightarrow\frac{-2}{\text{a}}\times3=-1$

$\Rightarrow\text{a}=6$

1. 1

Solution:

Let (x₁, y₁) be the required point.

The slope of the given line is m.

we have

$\text{y}^2=4\text{x}$

$\Rightarrow2\text{y}\frac{\text{dy}}{\text{dx}}=4$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{4}{2\text{y}}=\frac{2}{\text{y}}$

Slope of the tangent $=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(\text{x}_1,\text{y}_1)}=\frac{2}{\text{y}_1}$

Given:

Slope of tangent = m

Now,

$\frac{2}{\text{y}_1}=\text{m}\ ...(1)$

Because the given line is a tangent to the given curve at point (x₁, y₁) this point lies on both the line and the curve.

$\therefore\text{y}_1=\text{mx}+1$ and $\text{y}_1^2=4\text{x}_1$

$\Rightarrow\text{x}_1=\frac{\text{y}_1-1}{\text{m}}$ and $\text{x}_1=\frac{\text{y}_1^2}{4}$

So,

$\frac{\text{y}_1-1}{\text{m}}=\frac{\text{y}_1^2}{4}$

$\Rightarrow\frac{\text{y}_1-1}{\Big(\frac{2}{\text{y}_1}\Big)}=\frac{\text{y}_1{^2}}{4}\Big[\text{From} (1)$

$\Rightarrow\frac{\text{y}_1(\text{y}_1-1)}{2}=\frac{\text{y}_1^2}{4}$

$\Rightarrow2\text{y}_1^2-2\text{y}_1=\text{y}_1^2$

$\Rightarrow\text{y}_1^2-2\text{y}_1=0$

$\text{y}_1(\text{y}_1-2)=0$

$\Rightarrow\text{y}_1=0,2$

So, for $\text{y}_1=0,\text{m}=\frac{2}{0}=\infty$

For $\text{y}_1=2,\text{m}=\frac{2}{2}=1$

2. $(3,9)$

Solution:

Let (x₁, y₁) be the point where the given curve intersect the given line at the given angle.

Since, the point lie on the curve.

Hence, $\text{y}_1=6\text{x}_1-\text{x}_1^2$

Now, $\text{y}=6\text{x}-\text{x}^2$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=6-2\text{x}$

$\Rightarrow\text{m}_ 1=6-2\text{x}_1$

and

x + y = 0

$\Rightarrow1+\frac{\text{dy}}{\text{dx}}=0$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=-1$

$\Rightarrow\text{m}_2=-1$

it is given that the angles between them is $\frac{\pi}{4}$

$\therefore\tan\theta=\Big|\frac{\text{m}_1-\text{m}_2}{1+\text{m}_1\text{m}_2}\Big|$

$\Rightarrow\tan\frac{\pi}{4}=\Big|\frac{6-2\text{x}_1+1}{1-6+2\text{x}_1}\Big|$

$\Rightarrow1=\Big|\frac{7-2\text{x}_1}{2\text{x}_1-5}\Big|$

$\Rightarrow\Big|\frac{7-2\text{x}_1}{2\text{x}_1-5}\Big|=1\text{ or }\Big|\frac{7-2\text{x}_1}{2\text{x}_1-5}\Big|=-1$

$\Rightarrow7-2\text{x}_1=2\text{x}_1-5\text{ or }7-2\text{x}_1=-2\text{x}_1+5$

$\Rightarrow4\text{x}_1=12\text{ or }2=0$

$\Rightarrow\text{x}_1=3$

and

$\text{y}_1=6\text{x}_1-\text{x}_1^2=18-9=9$

$\therefore(\text{x}_1,\text{y}_1)=(3,9)$

4. None of these.

Solution:

$\text{y}=12\text{x}-\text{x}^2$

Slope of the tangent = 0

$\frac{\text{dy}}{\text{dx}}=0$

$12-2\text{x}=0$

$\Rightarrow\text{x}=6$

$\Rightarrow\text{y}=36$

Point on corve is (6, 36).

3. $\frac{\pi}{3}$

​​​​​​​Solution:

$\text{y}=2\sin^2\text{x}\Rightarrow\frac{\text{dy}}{\text{dx}}=4\sin\text{x}\cos\text{x}$

$\Rightarrow\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}-\frac{\pi}{6}}=\sqrt{3}=\text{m}_1$

$\text{y}=\cos2\text{x}\Rightarrow\frac{\text{dy}}{\text{dx}}=-2\sin2\text{x}$

$\Rightarrow\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}-\frac{\pi}{6}}=-\sqrt{3}=\text{m}_1$

$\tan\theta\Big|\frac{\sqrt{3}+\sqrt{3}}{1-3}\Big|=\sqrt{3}$

$\Rightarrow\theta=\tan^{-1}\sqrt{3}=\frac{\pi}{3}$

2. 0

Solution:

$\text{x}=3\text{t}^2+1$ and $\text{y}=\text{t}^3-1$

$\frac{\text{dx}}{\text{dt}}=6\text{t}$ and $\frac{\text{dy}}{\text{dt}}=3\text{t}^2$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{\text{t}}{2} \ ...(1)$ 

But,

$\text{x}=1$

$\Rightarrow3\text{t}^2+1=1$

$\Rightarrow\text{t}=0$

$\frac{\text{dy}}{\text{dx}}=\frac{\text{t}}{2}=0 \ (\because\text{From(i))}$

2. x + y - 1 = 0 = x - y - 2

Solution:

$\text{y}=\text{x}^2-3\text{x}+2$

Slope of tangent

$\frac{\text{dy}}{\text{dx}}=2\text{x}-3$

Tangent meets at x-axis hence y = 0.

$\text{x}^2-3\text{x}+2=0$

$(\text{x}-2)(\text{x}-1)=0$

$\text{x}=2$ or $\text{x}=1 $

For $\text{x}=2\Rightarrow\text{m}=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(2,0)}=1$

Equation of tangent m = -1, point (1, 0)

$\text{y}-0-1(\text{x}-1)$

$\Rightarrow\text{x}+\text{y}-1=0$

3. (0, 0)

Solution:

x = at² and y = 2at

$\Rightarrow\frac{\text{dx}}{\text{dt}}=2\text{at}$ and $\frac{\text{dy}}{\text{dt}}=2\text{a}$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2\text{a}}{2\text{at}}=\frac{2\text{a}}{\text{y}}$

Slope of tangent $=\frac{2\text{a}}{\text{y}}$

Tangent is perpendicular to y-axis.

$\Rightarrow$Tangent is parallel to x-axis

$\frac{2\text{a}}{\text{y}}=0$

$\text{a}=0$

$\Rightarrow\text{x}=0$ and $\text{y}=0$

Point is (0, 0).

1. x + y = 3

Solution:

x² = 4y

$2\text{x}=4\frac{\text{dy}}{\text{dx}}$

$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{2}$

$\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(1,2)}=\frac{1}{2}=\text{m}$

Slope of normal $=\frac{-1}{\frac{\text{dy}}{\text{dx}}}=\frac{-1}{\frac{\text{x}}{\text{2}}}=\frac{-2}{\text{x}}$

Let (X, Y) be the point where normal and curve intersect

$\therefore $ Slope of normal at (X, Y) $=\frac{-2}{\text{X}}$

Equation of normal passing through (X, Y) with slope $\frac{-2}{\text{X}}$ is

$\text{y}-\text{y}_1=\text{m}(\text{x}-\text{x}_1)$

$\text{y}-\text{Y}=\frac{-2}{\text{h}}(\text{x}-\text{X})$

Since normal passes through (1, 2) it will satisfy its equation

$\text{Y}=2+\frac{-2}{\text{h}}(\text{1}-\text{X})\ \dots(1)$

Since (X, Y) lies on the curve x² = 4y

X² = 4Y .......(2)

Using (1) and (2)

$\Rightarrow\ 2+\frac{-2}{\text{h}}(\text{1}-\text{X})=\frac{\text{X}^2}{4}$

$2+\frac{2}{\text{X}}-2=\frac{\text{X}^2}{4}$

$\frac{2}{\text{X}}=\frac{\text{X}^2}{4}$

$\text{X}^3=8$

$\text{X}=2$

Putting X = 2 in (2)

$\text{Y}=\frac{\text{X}^2}{4}=\frac{(2)^2}{4}=\frac{4}{4}=1$

Hence, X = 2, Y = 1

$\text{y}-\text{Y}=\frac{-2(\text{x}-\text{X})}{\text{X}}$

$\text{y}-\text{1}=\frac{-2(\text{x}-\text{2})}{\text{2}}$

$\text{y}-1=-1(\text{x}-2)$

$\text{y}-1=-\text{x}+2$

$\text{x}+\text{y}=2+1$

$\text{x}+\text{y}=3$

2. $\tan^{-1}\frac{3}{4}$

Solution:

Given:

$\text{y}^2=\text{x} \ ...(1)$

$\text{x}^2=\text{y} \ ...(2)$

Point = (1, 1)

On diffierentiating (1) w.r.t. x, we get

$2\text{y}\frac{\text{dy}}{\text{dx}}=1$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{2\text{y}}$

$\Rightarrow\text{m}_1=\frac{1}{2}$

On differentiating (2) w.r.t. x, we get

$2\text{x}=\frac{\text{dy}}{\text{dx}}$

$\Rightarrow\text{m}_2(1)=2$

Now,

$\tan\theta=\Big|\frac{\text{m}_1-\text{m}_2}{1+\text{m}_1\text{m}_2}\Big|=\Big|\frac{\frac{1}{2}-2}{1+\frac{1}{2}\times2}\Big|=\frac{3}{4}$

$\Rightarrow\theta=\tan^{-1}\big(\frac{3}{4}\big)$

3. $\frac{\pi}{2}$

Solution:

Given:

y² = 4ax ...(1)

x² = 4ay ...(2)

Point = (0, 0)

On differentiating (1) w.r.t. x,

$2\text{y}\frac{\text{dy}}{\text{dx}}=4\text{a}$

$\Rightarrow\frac{\text{dy}}{\text{dx}}\approx\frac{2\text{a}}{\text{y}}$

$\Rightarrow\text{m}_1\approx\infty$

Now, on differentiating (2) w.r.t. x,

$2\text{x}\frac{\text{dy}}{\text{dx}}=4\text{a}$

$\Rightarrow\frac{\text{dy}}{\text{dx}}\approx\frac{\text{x}}{2\text{u}}=0$

$\therefore\tan\theta=\Big|\frac{\text{m}_1-\text{m}_2}{1+\text{m}_1\text{m}_2}\Big|=\Big|\frac{\infty}{1+0}\Big|=\infty$

$\Rightarrow\theta=\tan^{-1}\infty=\frac{\pi}{2}$

2. $\frac{6}{7}$

Solution:

x = t² + 3t - 8, y = 2t² - 2t - 5

$\frac{\text{dx}}{\text{dt}}=2\text{t}+3,\frac{\text{dy}}{\text{dt}}=4\text{t}-2$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}$

$\frac{\text{dy}}{\text{dx}}=\frac{4\text{t}-2}{2\text{t}-3}$

Point is given by (2, -1)

x = 2 and y = -1

t² + 3t - 8 = 2 and 2t² - 2t - 5 = -1

t² + 3t - 10 = 0 and 2t² - 2t - 4 = 0

(t + 5) (t - 2) = 0 and (2 + 2) (t + 2) = 0

For t = 2 both equation has common solution.

⇒ Slope of the tangent $=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{t}-2}=\frac{4\times2-2}{2\times2+3}=\frac{6}{7}$

3. ab = 1

Solution:

Given:

$\text{y}=\text{ae}^\text{x} \ ...(1)$

$\text{y}=\text{be}^{\text{-x}} \ ...(2)$

Let the point of intersection of these two curves be (x₁, y₁).

Now,

On differentiating (1) w.r.t. x, we get

$\frac{\text{dy}}{\text{dx}}=\text{ae}{^\text{x}}$

$=\text{m}_1=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(\text{x}_1,\text{y}_1)}=\text{ae}^{\text{x}^{1}}$

Again, on diffirentiating (2)w.r.t. x, we get

$\frac{\text{dy}}{\text{dx}}=-\text{be}^{\text{-x}}$

$\Rightarrow\text{m}_2=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(\text{x}_1,\text{y}_1)}=-\text{be}^{-\text{x}_1}$

It is given that the curves cut orthaonally.

$\therefore\text{m}_1\times\text{m}_2=-1$

$\Rightarrow\text{ae}^{\text{x}_1}\times (-\text{be}^-\text{x}_1)=-1$

$\Rightarrow\text{ab}=1$

2. b = -1, c = 1

Solution:

We can find the slope of the line by differentiating w.r.t.x,

Slope of the given line = 1

Now,

$\text{y}=\text{x}^2+\text{bx}+\text{c} \ ...(1)$

Slope of the tangent $=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(1,1)}=2+\text{b}$

Given:

Slope of the tangent = 1

$\Rightarrow2+\text{b}=1$

$\text{b}=-1$

On substituting b = -1, x = 1 and y = 1 in (1), we get 

$\Rightarrow1=1-1+\text{c}$

$\text{c}=1$

1. x - 2y = 2

Solution:

Here,

$\text{y}=\text{x}(2-\text{x})=2\text{x}-\text{x}^2$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=2-2\text{x}$

Slope of the tangent $=\bigg(\frac{\text{dy}}{\text{dx}}\bigg)_{(2,0)}=2-4=-2$

Slope of the normal $\text{m}=\frac{-1}{-2}=\frac{1}{2}$

Given:

$(\text{x}_1,\text{y}_1)=(2,0)$

$\therefore$ Equation of the normal

$=\text{y}-\text{y}_1=\text{m}(\text{x}-\text{x}_1)$

$\Rightarrow\text{y}-0=\frac{1}{2}(\text{x}-2)$

$\Rightarrow2\text{y}=\text{x}-2$

$\Rightarrow\text{x}-2\text{y}=2$

1. $\frac{1}{2}$

Solution:

 = 2e^x and y = ae^−x

^{$\Big(\frac{\text{dy}}{\text{dx}}\Big)_1=2\text{e}^\text{x}$} and ^{$\Big(\frac{\text{dy}}{\text{dx}}\Big)_2=-\text{ae}^\text{x}$}

Given that $\Big(\frac{\text{dy}}{\text{dx}}\Big)_1\times\Big(\frac{\text{dy}}{\text{dx}}\Big)_2=-1$

$2\text{e}^\text{x}-\text{ae}^{-\text{x}}=-1$

$2\text{a}=1$

$\text{a}=\frac{1}{2}$​​​​​​​

3. c = y

Solution:

$\text{x}=\text{a}\cos^3\theta $ and $\text{y}=\text{a}\sin^3\theta$

$\text{x}=\text{a}\cos^3\frac{\pi}{4}$ and $\text{y}=\text{a}\sin^3\frac{\pi}{4}$

$\text{x}=\frac{\text{a}}{2\sqrt{2}}$ and $\frac{\text{a}}{2\sqrt{2}}$

Point is $\Big(\frac{\text{a}}{2\sqrt{2}},\frac{\text{a}}{2\sqrt{2}}\Big)$

$\text{x}=\text{a}\cos^3\theta$ and $\text{y}=\text{a}\sin^3\theta$

$\Rightarrow\cos\theta=3\sqrt{\frac{\text{x}}{\text{a}}}$ and $\sin\theta=3\sqrt{\frac{\text{y}}{\text{a}}}$

$\sin^2\theta+\cos^2\theta=1$

$\Rightarrow\Big(\frac{\text{x}}{\text{a}}\Big)^{\frac{2}{3}}+\Big(\frac{\text{y}}{\text{a}}\Big)^{\frac{2}{3}}=1$

3. x + y = 0

​​​​​​​Solution:

Given:

$\text{y}=\sin\text{x}$

On differentiating both sides w.r.t.x, we get

$\frac{\text{dy}}{\text{dx}}=\cos\text{x}$

Slope of the tangent $=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(0,0)}=\cos=1$

Slope of the normal, $\text{m}=\frac{-1}{1}=-1$

Given:

(x₁, y₁) = (0, 0)

$\therefore$ equation of the normal

= y - y₁ = m (x₁, y₁)

⇒ y - 0 = -1 (x - 0)

⇒ y = -x

⇒ x + y = 0

3. $\text{x}+\pi=0$

Solution:

Given:

$\text{y}=\sin\text{x}$

On differentinating both sides w.r.t. x,

We get,

$\frac{\text{dy}}{\text{dx}}=\cos\text{x}$

Slope of the tangent $=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(0,0)}=\cos0=1$

Slope of the normal $\text{m,}=\frac{-1}{1}=-1$

Given:

$(\text{x}_1,\text{y}_1)=(0,0)$

$\therefore$ Equation of the normal

$=\text{y}-\text{y}_1=\text{m}(\text{x}-\text{x}_1)$

$\Rightarrow\text{y}-0=-1(\text{x}-0)$

$\Rightarrow\text{y}=-\text{x}$

$\Rightarrow\text{x}+\text{y}=0$