EXERCISE 6.3

Question 1: Find the slope of the tangent to the curve $y=3 x^4-4 x$ at $x=4$ .

Solution: The slope of tangent to the curve $y=3 x^4-4 x$

$\frac{d y}{d x}=12 x^3-4$

If $x=4$ , then slope

$\Rightarrow \left[\frac{d y}{dx}\right]_{x=4}=12(4)^3-4=768-4=764$

Question 2: Find the slope of the tangent to the curve $y=\frac{x-1}{x-2}, x \neq 2$ at $x=10$ .

Solution: The slope of tangent to the curve

$y=\frac{x-1}{x-2}$

$\frac{d y}{d x}=\frac{(x-2) \cdot 1-(x-1) \cdot 1}{(x-2)^2}$

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

If $x=10$ , then slope $\left.=\frac{d y}{d x}\right]_{x=10}=\frac{-1}{(10-2)^2}=-\frac{1}{64}$

Question 3: Find the slope of the tangent to curve $y=x^3-x+1$ at the point whose $x$ -coordinate is 2 .

Solution: The slope of tangent to the curve

$y=x^3-x+1$

$\frac{d y}{d x}=3 x^2-1$

If $x$ -coordinate is 2 , then slope $\left.=\frac{d y}{d x}\right]_{x=2}$

$=3(2)^2-1=12-1=11$

Question 4: Find the slope of the tangent to the curve $y=x^3-3 x+2$ at the point whose $x$ -coordinate is 3 .

Solution: The slope of tangent to the curve

$y=x^3-3 x+2$

$\frac{d y}{d x}=3 x^2-3$

If $x$ -coordinate is 3 , then slope $\left.=\frac{d y}{d x}\right]_{x=3}=3(3)^2-3$

$=27-3=24$

Question 5: Find the slope of the normal to the curve $x=a \cos ^3 \theta, y=a \sin ^3 \theta$ at $\theta=\frac{\pi}{4}$ .

Solution: Here, $x=a \cos ^3 \theta$

$\Rightarrow \frac{d x}{d \theta}=3 a \cos ^2 \theta(-\sin \theta)$

$=-3 a \sin \theta \cos ^2 \theta$

and

$y=a \sin ^3 \theta$

$\Rightarrow \frac{d y}{d \theta}=3 a \sin ^2 \theta(\cos \theta)$

$=3 a \cos \theta \sin ^2 \theta$

The slope of normal to the curve

$-\frac{d x}{d y}=-\frac{\frac{d x}{d \theta}}{\frac{d y}{d \theta}}$

$=-\frac{-3 a \sin \theta \cos ^2 \theta}{3 a \cos \theta \sin ^2 \theta}=\cot \theta$

If $\theta=\frac{\pi}{4}$ , then slope of normal

$\left.\frac{d x}{d y}\right]_{\theta=\frac{\pi}{4}}=\cot \frac{\pi}{4}=1$

Question 6: Find the slope of the normal to the curve $x=1-a \sin \theta, y=b \cos ^2 \theta$ at $\theta=\frac{\pi}{2}$ .

Solution: Here, $x=1-a \sin \theta$

$\Rightarrow \frac{d x}{d \theta}$

$=-a \cos \theta$

and $y=b \cos ^2 \theta$

$\Rightarrow \frac{d y}{d \theta}=2 b \cos \theta(-\sin \theta)$

$=-2 b \cos \theta \sin \theta$

$\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}$

$=\frac{-2 b \cos \theta \sin \theta}{-a \cos \theta}$

$=\frac{2 b \sin \theta}{a}$

The slope of normal $=\frac{-1}{dy/dx}$

$= -\frac{a}{2 b \sin \theta}$

If $\theta=\frac{\pi}{2}$ , then the slope of normal

$=-\frac{a}{2 b \sin \frac{\pi}{2}}=-\frac{a}{2 b}$

Question 7: Find points at which the tangent to the curve $y=x^3-3 x^2-9 x+7$ is parallel to the $x$ -axis.

Solution: The slope of tangent to the curve

$y=x^3-3 x^2-9 x+7$

$\frac{d y}{d x}=3 x^2-6 x-9$

If the tangent is parallel to $x$ – axis, then slope

$\frac{d y}{d x}=0$

$\Rightarrow 3 x^2-6 x-9=0$

$\Rightarrow 3\left(x^2-2 x-3\right)=0$

$\Rightarrow 3(x-3)(x+1)=0$

$\Rightarrow x=-1,3$

If $x=-1$ , then

$y=(-1)^3-3(-1)^2-9(-1)+7$

$=-1-3+9+7=12$ ,

so, the points $=(-1,12)$

If $x=3$ , then

$y=(3)^3-3(3)^2-9(3)+7$

$=27-27-27+7=-20$ ,

so, the point $=(3,-20)$

Hence, on the points $(-1,12)$ and $(3,-20)$ , the tangents are parallel to $x$ – axis.

Question 8: Find a point on the curve $y=(x-2)^2$ at which the tangent is parallel to the chord joining the points $(2,0)$ and $(4,4)$ .

Solution: The slope of tangent to the curve

$y=(x-2)^2$

$\Rightarrow \frac{d y}{d x}=2(x-2)$

The slope of line joining the points $(2,0)$ and $(4,4)=\frac{4-0}{4-2}=2$

Given: The slope of tangent to the curve $=$ slope of line joining the points $(2,0)$ and $(4,4)$ .

$\Rightarrow 2(x-2)=2$

$\Rightarrow x-2=1 \quad \Rightarrow x=3$ ,

If $x=3$ , then $y=(3-2)^2=1$ ,

therefore, the required point $=(3,1)$

Question 9: Find the point on the curve $y=x^3-11 x+5$ at which the tangent is $y=x-11$ .

Solution: Given $y=x^3-11 x+5$

Slope of tangent to the curve

$\frac{d y}{d x}=3 x^2-11$

Slope of the line $y=x-11$

$\frac{dy}{dx}= 1$ .

Given that: The slope of tangent to the curve = the slope of the line

$\Rightarrow 2 x^2-11=1$

$\Rightarrow 3 x^2=12$

$\Rightarrow x^2=4$

$\Rightarrow x=\pm 2$

If $x=2$ , then

$y=(2)^3-11(2)+5=-9$ ,

therefore, the point $=(2,-9)$

If $x=-2$ , then

$y=(-2)^3-11(-2)+5=19$ ,

therefore, the point $=(-2,19)$

Out of the points $(2,-9)$ and $(-2,19)$ , only $(2,-9)$ satisfy the equation of line $y=x-11$ .

Hence, $(2,-9)$ is the point on $y=x^3-11 x+5$ at which tangent is $y=x-11$ .

Question 10: Find the equation of all lines having slope $-1$ that are tangents to the curve $y=\frac{1}{x-1}, x \neq 1$ .

Solution: curve $y=\frac{1}{x-1}$

The slope of tangent to the curve

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

Given that: The slope of tangent to curve $y=\frac{1}{x-1}$ is $-1$ .

$\Rightarrow-\frac{1}{(x-1)^2}=-1$

$\Rightarrow(x-1)^2=1$

$\Rightarrow x-1=\pm 1$

$\Rightarrow x=0,2$

If $x=0$ , then

$y=\frac{1}{0-1}=-1$ ,

therefore, the point $=(0,-1)$ .

Hence, the tangent: $y+1=-1(x-0)$

$\Rightarrow x+y+1=0$

If $x=2$ , then

$y=\frac{1}{2-1}=1$ ,

therefore, the point $=(2,1)$ .

Hence, the tangent: $y-1=-1(x-2)$

$\Rightarrow x+y-3=0$

Question 11: Find the equation of all lines having slope 2 which are tangents to the curve $y=\frac{1}{x-3}, x \neq 3$ .

Solution: curve $y=\frac{1}{x-3}$

The slope of tangent to the curve

$\frac{d y}{d x}=-\frac{1}{(x-3)^2}$

Given that: The slope of tangent to the curve $y=\frac{1}{x-3}$ is 2 .

$\Rightarrow-\frac{1}{(x-3)^2}=2$

$\Rightarrow(x-3)^2=-2$

The square of a real number can’t be negative. Hence, such tangents doesn’t exists.

Question 12: Find the equations of all lines having slope 0 which are tangent to the curve $y=\frac{1}{x^2-2 x+3}$ .

Solution: curve $y=\frac{1}{x^2-2 x+3}$

The slope of tangent to the curve

$\frac{d y}{d x}=-\frac{2 x-2}{\left(x^2-2 x+3\right)^2}$

Given that: The slope of tangent to the curve $y=\frac{1}{x^2-2 x+3}$ is 0 .

$\Rightarrow-\frac{2 x-2}{\left(x^2-2 x+3\right)^2}=0 \quad \Rightarrow 2 x-2=0 \quad \Rightarrow x=1$

If $x=1$ , then

$y=\frac{1}{1^2-2(1)+3}=\frac{1}{2^2}$ ,

therefore, the point $=\left(1, \frac{1}{2}\right)$ .

Hence, the tangent: $y-\frac{1}{2}=0(x-1)$

$\Rightarrow 2 y-1=0$

Question 13: Find points on the curve $\frac{x^2}{9}+\frac{y^2}{16}=1$ at which the tangents are

(i) parallel to $x$ -axis

(ii) parallel to $y$ -axis.

Solution: $\frac{x^2}{9}+\frac{y^2}{16}=1$

Differentiating with respect to $x$ ,

$\frac{2 x}{9}+\frac{2 y}{16} \frac{d y}{d x}=0$

$\Rightarrow \frac{d y}{d x}=-\frac{16 x}{9 y}$

The slope of tangent

$\frac{d y}{d x}=-\frac{16 x}{9 y}$

(i) If the tangent is parallel to $\mathrm{x}$ – axis, then slope of tangent

$\frac{d y}{d x}=0$

$\Rightarrow-\frac{16 x}{9 y}=0$

$\Rightarrow x=0$

Putting the value of $\mathrm{x}$ in $\frac{x^2}{9}+\frac{y^2}{16}=1$ ,

$\frac{y^2}{16}=1$

$\Rightarrow y^2=16$

$\Rightarrow y=\pm 4$

Therefore, the required points are $(0,4)$ and $(0,-4)$ .

(ii) If the tangent is parallel to $\mathrm{y}$ -axis, the slope of normal

$=-\frac{d x}{d y}=0$

$\Rightarrow \frac{9 y}{16 x}=0$

$\Rightarrow y=0$

Putting the value of $y$ in $\frac{x^2}{9}+\frac{y^2}{16}=1$ , we have

$\frac{x^2}{9}=1$

$\Rightarrow x^2=9$

$\Rightarrow x=\pm 3$

Therefore, the required points are $(3,0)$ and $(-3,0)$ .

Question 14: Find the equations of the tangent and normal to the given curves at the indicated points:

(i) $y=x^4-6 x^3+13 x^2-10 x+5$ at $(0,5)$

(ii) $y=x^4-6 x^3+13 x^2-10 x+5$ at $(1,3)$

(iii) $y=x^3$ at $(1,1)$

(iv) $y=x^2$ at $(0,0)$

(v) $x=\cos t, y=\sin t$ at $t=\frac{\pi}{4}$

Solution: (i) curve $y=x^4-6 x^3+13 x^2-10 x+5$

The slope of tangent to the curve

$\frac{d y}{d x}=4 x^3-18 x^2+26 x-10$

The slope of tangent at $(0,5)$ is

$\left.\frac{d y}{d x}\right]_{(0,5)}=4(0)^3-18(0)^2+26(0)-10=-10$

Therefore, the equation of tangent passing through $(0,5)$ is

$y-5=-10(x-0)$

$\Rightarrow 10 x+y=5$

The slope of normal at $(0,5)$ is

$\left.-\frac{d x}{d y}\right]_{(0,5)}=\frac{1}{10}$

Therefore, the equation of normal through $(0,5)$ is

$y-5=\frac{1}{10}(x-0)$

$\Rightarrow x-10 y+50=0$

(ii) curve $y=x^4-6 x^3+13 x^2-10 x+5$

The slope of tangent to the curve

$\frac{d y}{d x}=4 x^3-18 x^2+26 x-10$

The slope of tangent at $(1,3)$ is

$\left.\frac{d y}{d x}\right]_{(1,3)}=4(1)^3-18(1)^2+26(1)-10=2$

Therefore, the tangent at $(1,3)$ is $y-3=2(x-1) \Rightarrow y=2 x+1$

The slope of normal at $(1,3)$ is

$\left.-\frac{d x}{d y}\right]_{(1,3)}=-\frac{1}{2}$

The equation of normal through $(1,3)$ is

$y-3=-\frac{1}{2}(x-1)$

$\Rightarrow x+2 y=7$

(iii) curve $y=x^3$

The slope of tangent to the curve

$\frac{d y}{d x}=3 x^2$

The slope of tangent at $(1,1)$ is

$\left.\frac{d y}{d x}\right]_{(1,1)}=3(1)^2=3$

Therefore, the equation of tangent through $(1,1)$ is

$y-1=3(x-1) \Rightarrow y=3 x-2$

Slope of normal at $(1,1)$ is

$\left.-\frac{d x}{d y}\right]_{(1,1)}=-\frac{1}{3}$

Therefore, the equation of normal at $(1,1)$ is

$y-1=-\frac{1}{3}(x-1)$

$\Rightarrow x+3 y=4$

(iv) curve $y=x^2$

The slope of tangent to the curve

$\frac{d y}{d x}=2 x$

The slope of tangent at $(0,0)$ is

$\left.\frac{d y}{d x}\right]_{(0,0)}=0$

Therefore, the equation of tangent through $(0,0)$ is

$y-0=0(x-0)$

$\Rightarrow y=0$

The slope of normal at $(0,0)$ is

$\left.-\frac{d x}{d y}\right]_{(0,0)}=\frac{1}{0}$

Therefore, the equation of normal at $(0,0)$ is

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

$\Rightarrow x=0$

(v) curve $x=\cos t, y=\sin t$

The slope of tangent

$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$

$=\frac{\cos t}{-\sin t}=-\cot t$

The slope of tangent at $t=\frac{\pi}{4}$ is

$\left.\frac{d y}{d x}\right]_{t=\frac{\pi}{4}}=-1$

Therefore, the equation of tangent at $t=\frac{\pi}{4}$ is

$y-\sin \frac{\pi}{4}=-1\left(x-\cos \frac{\pi}{4}\right)$

$\Rightarrow y-\frac{1}{\sqrt{2}}=-1\left(x-\frac{1}{\sqrt{2}}\right)$

$\Rightarrow x+y=\sqrt{2}$

The slope of normal at $t=\frac{\pi}{4}$ is $\left.-\frac{d x}{d y}\right]_{t=\frac{\pi}{4}}=1$

Therefore, the equation of normal at $t=\frac{\pi}{4}$ is

$y-\frac{1}{\sqrt{2}}=1\left(x-\frac{1}{\sqrt{2}}\right)$

$\Rightarrow x=y$

Question 15: Find the equation of the tangent line to the curve $y=x^2-2 x+7$ which is

(a) parallel to the line $2 x-y+9=0$

(b) perpendicular to the line $5 y-15 x=13$ .

Solution: curve $y=x^2-2 x+7$

The slope of tangent

$\frac{d y}{d x}=2 x-2$

(i) The slope of the line $2 x-y+9=0$

$2-\frac{d y}{d x}=0$

$\Rightarrow \frac{dy}{dx}= 2$

If the tangent to $y=x^2-2 x+7$ is parallel to $2 x-y+9=0$ , then the slopes are equal

$2 x-2= 2 \Rightarrow x=2$

Putting $\mathrm{x}=2$ in curve $y=x^2-2 x+7$ ,

we have, $y=(2)^2-2(2)+7=7$ .

So, the point $=(2,7)$

Therefore, the equation of tangent at $(2,7)$ is given by

$y-7=2(x-2)$

$\Rightarrow 2 x-y+3=0$

(ii) The line $5 y-15 x=13$

$5\frac{d y}{d x}-15=0$

Slope the line

$\Rightarrow \frac{dy}{dx}= 3$

If the tangent of $y=x^2-2 x+7$ is perpendicular to $5 y-15 x=13$ , then

we have $(2 x-2) \times 3=-1$

$\Rightarrow x=\frac{5}{6}$

Putting $x=\frac{5}{6}$ in $y=x^2-2 x+7$ ,

we have $y=\left(\frac{5}{6}\right)^2-2\left(\frac{5}{6}\right)+7$

$=\frac{25-60+252}{36}=\frac{217}{36}$ ,

Therefore, the point $=\left(\frac{5}{6}, \frac{217}{36}\right)$

the slope of tangent $\left.\frac{d y}{d x}\right]_{\left(\frac{5}{6^{\prime}}, 36\right.}=2 \times \frac{5}{6}-2$

$=-\frac{1}{3}$

The equation of tangent at $\left(\frac{5}{6}, \frac{217}{36}\right)$ is given by

$y-\frac{217}{36}=-\frac{1}{3}\left(x-\frac{5}{6}\right)$

$\Rightarrow 12 x+36 y-227=0$

Question 16: Show that the tangents to the curve $y=7 x^3+11$ at the points where $x=2$ and $x=-2$ are parallel.

Solution: curve $y=7 x^3+11$

The slope of tangent

$\frac{d y}{d x}=21 x^2$

The slope of tangent at $x=2$

$\left.\frac{d y}{d x}\right]_{x=2}=21(2)^2=84$

The slope of tangent at $x=-2$

$\left.\frac{d y}{d x}\right]_{x=-2}=21(-2)^2=84$

The slope of tangent at $x=2$ and $x=-2$ is same.

Hence, the tangents are parallel to each other.

Question 17: Find the points on the curve $y=x^3$ at which the slope of the tangent is equal to the $y$ -coordinate of the point.

Solution : The curve $y=x^3$ is $\frac{d y}{d x}=3 x^2$

The slope of tangent

Let $\left(x_1, y_1\right)$ be any point on $y=x^3$ ,

$y_1 = x^3$ –(i)

where the slope of tangent is equal to $y$ – coordinate of the point.

$\left.y_1=\frac{d y}{d x}\right]_{\left(x_1, y_1\right)}=3 x_1^2$

$\Rightarrow y_1=3 x_1^2$ –(ii)

Solving the equation (i) and (ii), we have,

$x_1{ }^3=3 x_1{ }^2$

$\Rightarrow x_1^3-3 x_1^2=0$

$\Rightarrow x_1^2\left(x_1-3\right)=0$

$\Rightarrow x_1=0,3$

From the equation (1), we have

If $x_1=0$ , then $y_1=0$ ,

therefore, the required point $=(0,0)$

If $x_1=3$ , then $y_1=3(3)^2=27$ ,

therefore, the required point $=(3,27)$

Question 18: For the curve $y=4 x^3-2 x^5$ , find all the points at which the tangent passes through the origin.

Solution: The curve $y=4 x^3-2 x^5$

The slope of tangent

$\frac{d y}{d x}=12 x^2-10 x^4$

Let, the point $\left(x_1, y_1\right)$ lies on the curve $y=4 x^3-2 x^5$ ,

$y_1 = 4x_1^3-2x_1^5$ —(i)

The slope of tangent at $\left(x_1, y_1\right)$

$\left.\frac{d y}{d x}\right]_{\left(x_1, y_1\right)}=12 x_1{ }^2-10 x_1{ }^4$ —(ii)

Slope of the line through $(0,0)$ and $\left(x_1, y_1\right)$ is given by $=\frac{y_1-0}{x_1-0}=\frac{y_1}{x_1}$ –(iii)

From the equation (ii) and (iii), we have

$\frac{y_1}{x_1}=12 x_1{ }^2-10 x_1{ }^4$

$\Rightarrow y_1=12 x_1{ }^3-10 x_1{ }^5$ —(iv)

Solving the equation (1) and (iv), we have

$12 x_1^3-10 x_1^5=4 x_1^3-2 x_1^5$

$\Rightarrow 8 x_1{ }^3-8 x_1{ }^5=0$

$\Rightarrow x_1{ }^3\left(1-x_1^2\right)=0$

$\Rightarrow x_1=0, \pm 1$

From the equation (iv), we have

If $x_1=0$ , then

$y_1=12(0)^3-10(0)^5=0$ ,

therefore, the required point $=(0,0)$

If $x_1=1$ , then

$y_1=y_1=12(1)^3-10(1)^5=2$ ,

therefore, the required point $=(1,2)$

If $x_1=-1$ , then

$y_1=y_1=12(-1)^3-10(-1)^5=-2$ ,

therefore, the required point $=(-1,-2)$

Question 19: Find the points on the curve $x^2+y^2-2 x-3=0$ at which the tangents are parallel to the $x$ -axis.

Solution: $x^2+y^2-2 x-3=0$
Differentiating with respect to x:

$2 x+2 y \frac{d y}{d x}-2=0$

$\Rightarrow \frac{d y}{d x}=\frac{1-x}{y}$

The slope of tangent

$\frac{d y}{d x}=0$

The tangent is parallel to $x$ -axis.

Therefore,

$\frac{d y}{d x}=\frac{1-x}{y}=0$

$\Rightarrow 1-x=0$

$\Rightarrow x=1$ Putting in the curve

$x^2+y^2-2 x-3=0$ , we have

$1^2+y^2-2(1)-3=0$

$\Rightarrow y^2=4$

$\Rightarrow y=\pm 2 \text {, }$

Therefore, the points are $(1,2)$ and $(1,-2)$ .

The tangent at point $(1,2)$ is given by

$y-2=0(x-1) \Rightarrow y=2$

The tangent at point $(1,-2)$ is given by

$y-(-2)=0(x-1) \Rightarrow y=-2$

Question 20: Find the equation of the normal at the point $\left(a m^2, a m^3\right)$ for the curve $a y^2=x^3$ .

Solution : $a y^2=x^3$

Differentiating with respect to $x$ , we have

$2 a y \frac{d y}{d x}=3 x^2$

$\Rightarrow \frac{d y}{d x}=\frac{3 x^2}{2 a y}$

The slope of normal to the curve

$= \frac{-1}{dy/dx} = -\frac{d x}{d y}$

$=-\frac{2 a y}{3 x^2}$

The slope of normal at $\left(a m^2, a m^3\right)$ is given by

$=-\frac{2 a\left(a m^3\right)}{3\left(a m^2\right)^2}=-\frac{2}{3 m}$

The equation of normal at $\left(a m^2, a m^3\right)$

$y-a m^3=-\frac{2}{3 m}\left(x-a m^2\right)$

$\Rightarrow 2 x+3 m y=3 a m^4+2 a m^2$

Question 21: Find the equation of the normals to the curve $y=x 3+2 x+6$ which are parallel to the line $x+14 y+4=0$ .

Solution: $y=x^3+2 x+6$

Differentiating with respect to $x$ , we have

$\frac{d y}{d x}=3 x^2+2$

The slope of normal to is

$= \frac{-1}{dy/dx} = -\frac{d x}{d y}$

$=-\frac{1}{3 x^2+2}$

The slope of line $x+14 y+4=0$ is

$\frac{d y}{d x}=-\frac{1}{14}$

If the normal to $y=x^3+2 x+6$ is parallel to line $x+14 y+4=0$ , the both have the same slope.

Therefore, $-\frac{1}{3 x^2+2}=-\frac{1}{14}$

$\Rightarrow 3 x^2+2=14$

$\Rightarrow x^2=4$

$\Rightarrow x=\pm 2$

Putting $\mathrm{x}=2$ in $y=x^3+2 x+6$ , we have,

$y=(2)^3+2(2)+6=18$ ,

therefore the point $=(2,18)$

The equation of normal at $(2,18)$ is given by

$y-18=-\frac{1}{14}(x-2)$

$\Rightarrow x+14 y=254$

Putting $\mathrm{x}=-2$ in $y=x^3+2 x+6$ , we have,

$y=(-2)^3+2(-2)+6=-6$ ,

therefore, point $=(-2,-6)$

Therefore, the equation of normal at $(-2,-6)$ is given by

$y+6=-\frac{1}{14}(x+2)$

$\Rightarrow x+14 y+86=0$

Question 22: Find the equations of the tangent and normal to the parabola $y^2=4 a x$ at the point $\left(a t^2, 2 a t\right)$ .

Solution : $y^2=4 a x$

Differentiating with respect to $x$ , we have

$2 y \frac{d y}{d x}=4 a$

$\Rightarrow \frac{d y}{d x}=\frac{4 a}{2 y}$

The slope of normal to $y^2=4 a x$ is given by

$\frac{-1}{dy/dx} = -\frac{d x}{d y}$

$=-\frac{2 y}{4 a}$

Therefore, the slope of normal at $\left(a t^2, 2 a t\right)$

$=-\frac{2(2 a t)}{4 a}=-t$

The equation of normal at $\left(a t^2, 2 a t\right)$

$y-2 a t=-t\left(x-a t^2\right)$

$\Rightarrow t x+y=2 a t+a t^3$

Question 23: Prove that the curves $x=y 2$ and $x y=k$ cut at right angles if $8 k^2=1$ .

Solution: $x=y^2$ –(i)

$x y=k$
–(ii)

Putting the value of $x$

Solving (i) and (ii)

$y^3=k$

$\Rightarrow y=k^{\frac{1}{3}}$

Putting the value of $y$ in equation (1), we have, $x=k^{\frac{2}{3}}$

Hence, the two curves intersects at the point $\left(k^{\frac{2}{3}}, k^{\frac{1}{3}}\right)$ .

$x=y^2$

Differentiating with respect to $x$ , we have

$1=2 y \frac{d y}{d x}$

$\Rightarrow \frac{d y}{d x}=\frac{1}{2 y}$

The curve $x=y^2$

The slope of tangent

$\frac{d y}{d x}=\frac{1}{2 y}$

Slope of tangent at point $\left(k^{\frac{2}{3}}, k^{\frac{1}{3}}\right)$ is given by

$\left.m_1=\frac{d y}{d x}\right]_\left(k^{\frac{2}{3}}, k^{\frac{1}{3}}\right)=\frac{1}{2 k^{\frac{1}{3}}}$

Differentiating $x y=k$ with respect to $x$ , we have

$y+x \frac{d y}{d x}=0$

$\Rightarrow \frac{d y}{d x}=-\frac{y}{x}$

The slope of tangent to the curve $x y=k$ is $\frac{d y}{d x}=-\frac{y}{x}$

The slope of tangent at $\left(k^{\frac{2}{3}}, k^{\frac{1}{3}}\right)$ is given by

$\left.m_2=\frac{d y}{d x}\right]_{\left(k^{\frac{2}{3}, k^{\frac{1}{3}}}\right)}=-\frac{k^{\frac{1}{3}}}{k^{\frac{2}{3}}}$

$m_1 \times m_2=\frac{1}{2 k^{\frac{1}{3}}} \times\left(-\frac{k^{\frac{1}{3}}}{k^{\frac{2}{3}}}\right)$

$=-\frac{1}{2 k^{\frac{2}{3}}}$

$=-\frac{1}{8^{\frac{1}{3}} k^{\frac{2}{3}}}$

$=-\frac{1}{\left(8 k^2\right)^{\frac{1}{3}}}$

$=-\frac{1}{(1)^{\frac{1}{3}}}=-1 \quad\left[\because 8 k^2=1\right]$

$\Rightarrow$ The curves $x=y^2$ and $x y=k$ intersect each other at right angle.

Question 24: Find the equations of the tangent and normal to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ at the point $\left(x_0, y_0\right)$ .

Solution: Since $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Differentiating with respect to $x$ , we have

$\frac{2 x}{a^2}-\frac{2 y}{b^2} \frac{d y}{d x}=$

$\Rightarrow \frac{d y}{d x}=\frac{x}{\mathrm{y}} \cdot \frac{b^2}{a^2}$

The slope of tangent to hyperbola is

$\frac{d y}{d x}=\frac{x}{y} \cdot \frac{b^2}{a^2}$

Therefore, the slope of tangent at $\left(x_0, y_0\right)$ is given by

$\left.\frac{d y}{d x}\right]_{\left(x_0, y_0\right)}=\frac{d y}{d x}=\frac{x_0}{y_0} \cdot \frac{b^2}{a^2}$

The equation of tangent at $\left(x_0, y_0\right)$ :

$y-y_0=\frac{x_0}{y_0} \cdot \frac{b^2}{a^2}\left(x-x_0\right)$

$\Rightarrow a^2 y y_0-a^2 y_0^2=b^2 x x_0-b^2 x_0^2$

$\Rightarrow b^2 x_0^2-a^2 y_0^2=b^2 x x_0-a^2 y y_0$

$\Rightarrow \frac{x_0^2}{a^2}-\frac{y_0^2}{b^2}=\frac{x x_0}{b^2}-\frac{y y_0}{b^2} \quad\left[\text { Dividing by } a^2 b^2\right]$

$\Rightarrow 1=\frac{x x_0}{a^2}-\frac{y y_0}{b^2} \quad\left[\because \frac{x_0^2}{a^2}-\frac{y_0^2}{b^2}=1\right]$

The slope of normal to hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is given by

$=\frac{-1}{dy/dx} = -\frac{d x}{d y}=-\frac{d x}{d y}=-\frac{y}{x} \cdot \frac{a^2}{b^2}$

The slope of normal at $\left(x_0, y_0\right)$ is given by

$=-\frac{d x}{d y}=-\frac{y_0}{x_0} \cdot \frac{a^2}{b^2}$

The equation of normal at $\left(x_0, y_0\right)$ :

$y-y_0=-\frac{y_0}{x_0} \cdot \frac{a^2}{b^2}\left(x-x_0\right)$

$\Rightarrow \frac{y-y_0}{a^2 y_0}=-\frac{\left(x-x_0\right)}{b^2 x_0}$

$\Rightarrow \frac{\left(x-x_0\right)}{b^2 x_0}+\frac{y-y_0}{a^2 y_0}=0$

Question 25: Find the equation of the tangent to the curve $y=\sqrt{3 x-2}$ which is parallel to the line $4 x-2 y+5=0$ .

Solution: The curve $y=\sqrt{3 x-2}$

The slope of tangent is

$\frac{d y}{d x}=\frac{3}{2 \sqrt{3 x-2}}$

The slope of line $4 x-2 y+5=0$

$\frac{d y}{d x}=2$

If the tangent to $y=\sqrt{3 x-2}$ is parallel to $4 x-2 y+5=0$ , then their slopes will be equal.

Therefore, $\frac{3}{2 \sqrt{3 x-2}}=2$

$\Rightarrow \sqrt{3 x-2}=\frac{3}{4}$

$\Rightarrow 3 x-2=\frac{9}{16}$

$\Rightarrow 3 x=\frac{41}{16}$

$\Rightarrow x=\frac{41}{48}$

Putting $x=\frac{41}{48}$ in the equation $y=\sqrt{3 x-2}$ , we have,

$y=\sqrt{3 \times \frac{41}{48}-2}$

$=\sqrt{\frac{41}{16}-2}$

$=\sqrt{\frac{9}{16}}=\frac{3}{4}$

Therefore, the point $=\left(\frac{41}{48}, \frac{3}{4}\right)$

Therefore, the equation of tangent at $\left(\frac{41}{48}, \frac{3}{4}\right)$ :

$y-\frac{3}{4}=2\left(x-\frac{41}{48}\right)$

$\Rightarrow \frac{4 y-3}{4}=\frac{96 x-82}{48}$

$\Rightarrow 48 y-36=96 x-82$

$\Rightarrow 24 y-18=48 x-41$

$\Rightarrow 48 x-24 y=23$

Choose the correct answer in Exercises 26 and 27.

Question 26: The slope of the normal to the curve $y=2 x^2+3 \sin x$ at $x=0$ is

(A) 3

(B) $\frac{1}{3}$

(D) $-\frac{1}{3}$

Solution: Differentiating $y=2 x^2+3 \sin x$ with respect to $x$ , we have,

$\frac{d y}{d x}=4 x+3 \cos x$

Slope of normal to $y=2 x^2+3 \sin x$ is given by

$-\frac{d x}{d y}=-\frac{1}{4 x+3 \cos x}$
The equation of normal at $x=0:$

$=\frac{-1}{dy/dx}=-\left.\frac{d x}{d y}\right]_{x=0}=-\frac{1}{4 \times 0+3 \times 1}$

$=-\frac{1}{3}$

Hence, the option (D) is correct.

Question 27: The line $y=x+1$ is a tangent to the curve $y^2=4 x$ at the point

(A) $(1,2)$

(B) $(2,1)$

(D) $(-1,2)$

Solution: Differentiating $y^2=4 x$ with respect to $x$ , we have $2 y \frac{d y}{d x}=4$

The slope of tangent to the curve $y^2=4 x$ is $\frac{d y}{d x}=\frac{2}{y}$

Slope of the line $y=x+1$ is given by $\frac{d y}{d x}=1$

If the tangent of $y^2=4 x$ is $y=x+1$ .

Therefore,

$\frac{2}{y}=1 \quad \Rightarrow y=2$

Putting $y=2$ in $y^2=4 x$ , we have,

$(2)^2=4 x$

$\Rightarrow x=1$ .

Therefore, the point $=(1,2)$

Hence, the option (A) is correct.

Case study application of derivative 4 — These days chinese and Indian troops are engaged in aggressive melee, face-offs skirmishes

class 12 maths ex 6.3 ncert solution

EXERCISE 6.3

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