# NCERT Solutions for Exercise 6.3 Class 12 Maths Chapter 10 - Application of Derivatives

NCERT solutions for exercise 6.3 Class 12 Maths chapter 6 is related to the topic normal and tangents. The equations of normals and tangent to a curve and questions related to these are discussed in exercise 6.3 Class 12 Maths. There are twenty-seven questions presented in the Class 12 Maths chapter 6 exercise 6.3. These questions in Class 12th Maths chapter 6 exercise 6.3 are solved by our Mathematics subject matter experts and are reliable and according to the CBSE pattern. The NCERT solutions for Class 12 Maths chapter 6 exercise 6.3 along with all other exercises of the NCERT chapter applications of derivatives gives a good idea of concepts discussed in the NCERT Book.

• Application of Derivatives Exercise 6.1

• Application of Derivatives Exercise 6.2

• Application of Derivatives Exercise 6.4

• Application of Derivatives Exercise 6.5

• Application of Derivatives Miscellaneous Exercise

## Application of Derivatives Class 12 Chapter 6 Exercise 6.3

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

Given curve is,
$y = 3 x ^4 - 4x$
Now, the slope of the tangent at point x =4 is given by
$\left ( \frac{dy}{dx} \right )_{x=4} = 12x^3 - 4$
$= 12(4)^3-4$
$= 12(64)-4 = 768 - 4 =764$

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

Given curve is,

$y = \frac{x-1}{x-2}$
The slope of the tangent at x = 10 is given by
$\left ( \frac{dy}{dx} \right )_{x=10}= \frac{(1)(x-2)-(1)(x-1)}{(x-2)^2} = \frac{x-2-x+1}{(x-2)^2} = \frac{-1}{(x-2)^2}$
at x = 10
$= \frac{-1}{(10-2)^2} = \frac{-1}{8^2} = \frac{-1}{64}$
hence, slope of tangent at x = 10 is $\frac{-1}{64}$

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

Given curve is,
$y = x ^3 - x +1$
The slope of the tangent at x = 2 is given by
$\left ( \frac{dy}{dx} \right )_{x=2} = 3x^2 - 1 = 3(2)^2 - 1= 3\times 4 - 1 = 12 - 1 = 11$
Hence, the slope of the tangent at point x = 2 is 11

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

Given curve is,
$y = x ^3 - 3x +2$
The slope of the tangent at x = 3 is given by
$\left ( \frac{dy}{dx} \right )_{x=3} = 3x^2 - 3 = 3(3)^2 - 3= 3\times 9 - 3 = 27 - 3 = 24$
Hence, the slope of tangent at point x = 3 is 24

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

The slope of the tangent at a point on a given curve is given by
$\left ( \frac{dy}{dx} \right )$
Now,
$\left ( \frac{dx}{d \theta} \right )_{\theta=\frac{\pi}{4}} = 3a\cos^2 \theta(-\sin \theta) = 3a(\frac{1}{\sqrt2})^2(-\frac{1}{\sqrt2}) = -\frac{3\sqrt2 a}{4}$
Similarly,
$\left ( \frac{dy}{d \theta} \right )_{\theta=\frac{\pi}{4}} = 3a\sin^2 \theta(\cos \theta) = 3a(\frac{1}{\sqrt2})^2(\frac{1}{\sqrt2}) = \frac{3\sqrt2 a}{4}$
$\left ( \frac{dy}{dx} \right ) = \frac{\left ( \frac{dy}{d\theta} \right )}{\left ( \frac{dx}{d\theta} \right )} = \frac{\frac{3\sqrt2 a}{4}}{-\frac{3\sqrt2 a}{4}} = -1$
Hence, the slope of the tangent at $\theta = \frac{\pi}{4}$ is -1
Now,
Slope of normal = $-\frac{1}{slope \ of \ tangent}$ = $-\frac{1}{-1} = 1$
Hence, the slope of normal at $\theta = \frac{\pi}{4}$ is 1

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

The slope of the tangent at a point on given curves is given by
$\left ( \frac{dy}{dx} \right )$
Now,
$\left ( \frac{dx}{d \theta} \right )_{\theta=\frac{\pi}{2}} = -a(\cos \theta)$
Similarly,
$\left ( \frac{dy}{d \theta} \right )_{\theta=\frac{\pi}{2}} = 2b\cos \theta(-\sin \theta)$
$\left ( \frac{dy}{dx} \right )_{x=\frac{\pi}{2}} = \frac{\left ( \frac{dy}{d\theta} \right )}{\left ( \frac{dx}{d\theta} \right )} = \frac{-2b\cos \theta \sin \theta}{-a\cos \theta} = \frac{2b\sin \theta}{a} = \frac{2b\times1}{a} = \frac{2b}{a}$
Hence, the slope of the tangent at $\theta = \frac{\pi}{2}$ is $\frac{2b}{a}$
Now,
Slope of normal = $-\frac{1}{slope \ of \ tangent}$ = $-\frac{1}{\frac{2b}{a}} = -\frac{a}{2b}$
Hence, the slope of normal at $\theta = \frac{\pi}{2}$ is $-\frac{a}{2b}$

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

We are given :

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

Differentiating the equation with respect to x, we get :

$\frac{dy}{dx}\ =\ 3x^2\ -\ 6x\ -\ 9\ +\ 0$

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

or $\frac{dy}{dx}\ =\ 3\left ( x+1 \right )\left ( x-3 \right )$

It is given that tangent is parallel to the x-axis, so the slope of the tangent is equal to 0.

So,

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

or $0\ =\ 3\left ( x+1 \right )\left ( x-3 \right )$

Thus, Either x = -1 or x = 3

When x = -1 we get y = 12 and if x =3 we get y = -20

So the required points are (-1, 12) and (3, -20).

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).

Points joining the chord is (2,0) and (4,4)
Now, we know that the slope of the curve with given two points is
$m = \frac{y_2-y_1}{x_2 - x_1} = \frac{4-0}{4-2} = \frac{4}{2} =2$
As it is given that the tangent is parallel to the chord, so their slopes are equal
i.e. slope of the tangent = slope of the chord
Given the equation of the curve is $y = ( x-2)^2$
$\therefore \frac{dy}{dx} = 2(x-2) = 2$
$(x-2) = 1\\ x = 1+2\\ x=3$
Now, when $x=3$ $y=(3- 2)^2 = (1)^2 = 1$
Hence, the coordinates are (3, 1)

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

We know that the equation of a line is y = mx + c
Know the given equation of tangent is
y = x - 11
So, by comparing with the standard equation we can say that the slope of the tangent (m) = 1 and value of c if -11
As we know that slope of the tangent at a point on the given curve is given by $\frac{dy}{dx}$
Given the equation of curve is
$y = x^3 - 11x + 5$
$\frac{dy}{dx} = 3x^2 -11$
$3x^2 -11 = 1\\ 3x^2 = 12 \\ x^2 = 4 \\ x = \pm2$
When x = 2 , $y = 2^3 - 11(2) +5 = 8 - 22+5=-9$
and
When x = -2 , $y = (-2)^3 - 11(22) +5 = -8 + 22+5=19$
Hence, the coordinates are (2,-9) and (-2,19), here (-2,19) does not satisfy the equation y=x-11

Hence, the coordinate is (2,-9) at which the tangent is $y = x -11$

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

We know that the slope of the tangent of at the point of the given curve is given by $\frac{dy}{dx}$

Given the equation of curve is
$y = \frac{1}{x-1}$
$\frac{dy}{dx} = \frac{-1}{(1-x)^2}$
It is given thta slope is -1
So,
$\frac{-1}{(1-x)^2} = -1 \Rightarrow (1-x)^2 = 1 = 1 - x = \pm 1 \\ \\ x = 0 \ and \ x = 2$
Now, when x = 0 , $y = \frac{1}{x-1} = \frac{1}{0-1} = -1$
and
when x = 2 , $y = \frac{1}{x-1} = \frac{1}{(2-1)} = 1$
Hence, the coordinates are (0,-1) and (2,1)
Equation of line passing through (0,-1) and having slope = -1 is
y = mx + c
-1 = 0 X -1 + c
c = -1
Now equation of line is
y = -x -1
y + x + 1 = 0
Similarly, Equation of line passing through (2,1) and having slope = -1 is
y = mx + c
1 = -1 X 2 + c
c = 3
Now equation of line is
y = -x + 3
y + x - 3 = 0

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

We know that the slope of the tangent of at the point of the given curve is given by $\frac{dy}{dx}$

Given the equation of curve is
$y = \frac{1}{x-3}$
$\frac{dy}{dx} = \frac{-1}{(x-3)^2}$
It is given that slope is 2
So,
$\frac{-1}{(x-3)^2} = 2 \Rightarrow (x-3)^2 = \frac{-1}{2} = x-3 = \pm \frac{\sqrt-1}{\sqrt2} \\ \\$
So, this is not possible as our coordinates are imaginary numbers
Hence, no tangent is possible with slope 2 to the curve $y = \frac{1}{x-3}$

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

We know that the slope of the tangent at a point on the given curve is given by $\frac{dy}{dx}$

Given the equation of the curve as
$y = \frac{1}{x^2 - 2x + 3}$
$\frac{dy}{dx} = \frac{-(2x-2)}{(x^2-2x+3)^2}$
It is given thta slope is 0
So,
$\frac{-(2x-2)}{(x^2 - 2x +3)^2} = 0 \Rightarrow 2x-2 = 0 = x = 1$
Now, when x = 1 , $y = \frac{1}{x^2-2x+3} = \frac{1}{1^2-2(1)+3} = \frac{1}{1-2+3} =\frac{1}{2}$

Hence, the coordinates are $\left ( 1,\frac{1}{2} \right )$
Equation of line passing through $\left ( 1,\frac{1}{2} \right )$ and having slope = 0 is
y = mx + c
1/2 = 0 X 1 + c
c = 1/2
Now equation of line is
$y = \frac{1}{2}$

Question:13(i) Find points on the curve $\frac{x^2 }{9} + \frac{y^2 }{16} = 1$ at which the tangents are parallel to x-axis

Parallel to x-axis means slope of tangent is 0
We know that slope of tangent at a given point on the given curve is given by $\frac{dy}{dx}$
Given the equation of the curve is
$\frac{x^2 }{9} + \frac{y^2 }{16} = 1 \Rightarrow 9y^2 = 144(1-16x^2)$
$18y\frac{dy}{dx} = -32x$
$\frac{dy}{dx} = \frac{(-32x)}{18y} = 0 \Rightarrow x = 0$
From this, we can say that $x = 0$
Now. when $x = 0$ , $\frac{0^2 }{9} + \frac{y^2 }{16} = 1\Rightarrow \frac{y^2}{16} = 1 \Rightarrow y = \pm 4$
Hence, the coordinates are (0,4) and (0,-4)

Question:13(ii) Find points on the curve $\frac{x^2}{9} + \frac{y^2}{16} = 1$ at which the tangents are parallel to y-axis

Parallel to y-axis means the slope of the tangent is $\infty$ , means the slope of normal is 0
We know that slope of the tangent at a given point on the given curve is given by $\frac{dy}{dx}$
Given the equation of the curve is
$\frac{x^2 }{9} + \frac{y^2 }{16} = 1 \Rightarrow 9y^2 = 144(1-16x^2)$
$18y\frac{dy}{dx} = 144(1-32x)$
$\frac{dy}{dx} = \frac{-32x}{18y} = \infty$
Slope of normal = $-\frac{dx}{dy} = \frac{18y}{32x} = 0$
From this we can say that y = 0
Now. when y = 0, $\frac{x^2 }{9} + \frac{0^2 }{16} \Rightarrow 1 = x = \pm 3$
Hence, the coordinates are (3,0) and (-3,0)

Question:14(i) Find the equations of the tangent and normal to the given curves at the indicated
points:
$y = x^4 - 6x^3 + 13x^2 - 10x + 5 \: \: at\: \: (0, 5)$

We know that Slope of the tangent at a point on the given curve is given by $\frac{dy}{dx}$
Given the equation of the curve
$y = x^4 - 6x^3 + 13x^2 - 10x + 5$
$\frac{dy}{dx}= 4x^3 - 18x^2 + 26x- 10$
at point (0,5)
$\frac{dy}{dx}= 4(0)^3 - 18(0)^2 + 26(0) - 10 = -10$
Hence slope of tangent is -10
Now we know that,
$slope \ of \ normal = \frac{-1}{slope \ of \ tangent} = \frac{-1}{-10} = \frac{1}{10}$
Now, equation of tangent at point (0,5) with slope = -10 is
$y = mx + c\\ 5 = 0 + c\\ c = 5$
equation of tangent is
$y = -10x + 5\\ y + 10x = 5$
Similarly, the equation of normal at point (0,5) with slope = 1/10 is
$\\y = mx + c \\5 = 0 + c \\c = 5$
equation of normal is
$\\y = \frac{1}{10}x+5 \\ 10y - x = 50$

Question:14(ii) Find the equations of the tangent and normal to the given curves at the indicated
points:
$y = x^4 - 6x^3 + 13x^2 - 10x + 5 \: \: at \: \: (1, 3)$

We know that Slope of tangent at a point on given curve is given by $\frac{dy}{dx}$
Given equation of curve
$y = x^4 - 6x^3 + 13x^2 - 10x + 5$
$\frac{dy}{dx}= 4x^3 - 18x^2 + 26x - 10$
at point (1,3)
$\frac{dy}{dx}= 4(1)^3 - 18(1)^2 + 26(1) - 10 = 2$
Hence slope of tangent is 2
Now we know that,
$slope \ of \ normal = \frac{-1}{slope \ of \ tangent} = \frac{-1}{2}$
Now, equation of tangent at point (1,3) with slope = 2 is
y = 2x + 1
y -2x = 1
Similarly, equation of normal at point (1,3) with slope = -1/2 is
y = mx + c
$3 = \frac{-1}{2}\times 1+ c$
$c = \frac{7}{2}$
equation of normal is
$y = \frac{-1}{2}x+\frac{7}{2} \\ 2y + x = 7$

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

$y = x^3\: \: at \: \: (1, 1)$

We know that Slope of the tangent at a point on the given curve is given by $\frac{dy}{dx}$
Given the equation of the curve
$y = x^3$
$\frac{dy}{dx}= 3x^2$
at point (1,1)
$\frac{dy}{dx}= 3(1)^2 = 3$
Hence slope of tangent is 3
Now we know that,
$slope \ of \ normal = \frac{-1}{slope \ of \ tangent} = \frac{-1}{3}$
Now, equation of tangent at point (1,1) with slope = 3 is
$y = mx + c\\ 1 = 1 \times 3 + c\\ c = 1 - 3 = -2$
equation of tangent is
$y - 3x + 2 = 0$
Similarly, equation of normal at point (1,1) with slope = -1/3 is
y = mx + c
$1 = \frac{-1}{3}\times 1+ c$
$c = \frac{4}{3}$
equation of normal is
$y = \frac{-1}{3}x+\frac{4}{3} \\ 3y + x = 4$

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

$y = x^2\: \: at\: \: (0, 0)$

We know that Slope of the tangent at a point on the given curve is given by $\frac{dy}{dx}$
Given the equation of the curve
$y = x^2$
$\frac{dy}{dx}= 2x$
at point (0,0)
$\frac{dy}{dx}= 2(0)^2 = 0$
Hence slope of tangent is 0
Now we know that,
$slope \ of \ normal = \frac{-1}{slope \ of \ tangent} = \frac{-1}{0} = -\infty$
Now, equation of tangent at point (0,0) with slope = 0 is
y = 0
Similarly, equation of normal at point (0,0) with slope = $-\infty$ is

$\\y = x \times -\infty + 0\\ x = \frac{y}{-\infty}\\ x=0$

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

$x = \cos t , y = \sin t \: \: at \: \: t = \pi /4$

We know that Slope of the tangent at a point on the given curve is given by $\frac{dy}{dx}$
Given the equation of the curve
$x = \cos t , y = \sin t$
Now,
$\frac{dx}{dt} = -\sin t$ and $\frac{dy}{dt} = \cos t$
Now,
$\left ( \frac{dy}{dx} \right )_{t=\frac{\pi}{4}} = \frac{ \frac{dy}{dt}}{ \frac{dx}{dt}} = \frac{\cos t}{-\sin t} = -\cot t = =- \cot \frac{\pi}{4} = -1$
Hence slope of the tangent is -1
Now we know that,
$slope \ of \ normal = \frac{-1}{slope \ of \ tangent} = \frac{-1}{-1} = 1$
Now, the equation of the tangent at the point $t = \frac{\pi}{4}$ with slope = -1 is
$x= \cos \frac{\pi}{4} = \frac{1}{\sqrt2}$ and

$y= \sin \frac{\pi}{4} = \frac{1}{\sqrt2}$
equation of the tangent at

$t = \frac{\pi}{4}$ i.e. $\left ( \frac{1}{\sqrt2}, \frac{1}{\sqrt2}\right )$ is

$y- y_1 = m(x-x_1)\\ y-\frac{1}{\sqrt2} = -1(x- \frac{1}{\sqrt2})\\ \sqrt2y + \sqrt2x = 2\\ y + x = \sqrt2$
Similarly, the equation of normal at $t = \frac{\pi}{4}$ with slope = 1 is
$x= \cos \frac{\pi}{4} = \frac{1}{\sqrt2}$ and

$y= \sin \frac{\pi}{4} = \frac{1}{\sqrt2}$
equation of the tangent at

$t = \frac{\pi}{4}$ i.e. $\left ( \frac{1}{\sqrt2}, \frac{1}{\sqrt2}\right )$ is
$\\y- y_1 = m(x-x_1)\\ y-\frac{1}{\sqrt2} = 1(x- \frac{1}{\sqrt2})\\ \sqrt2y - \sqrt2x = 0\\ y - x = 0\\ x=y$

Question:15(a) Find the equation of the tangent line to the curve $y = x^2 - 2x +7$ which is parallel to the line $2x - y + 9 = 0$

Parellel to line $2x - y + 9 = 0$ means slope of tangent and slope of line is equal
We know that the equation of line is
y = mx + c
on comparing with the given equation we get slope of line m = 2 and c = 9
Now, we know that the slope of tangent at a given point to given curve is given by $\frac{dy}{dx}$
Given equation of curve is
$y = x^2 - 2x +7$
$\frac{dy}{dx} = 2x - 2 = 2\\ \\ x = 2$
Now, when x = 2 , $y = (2)^2 - 2(2) +7 =4 - 4 + 7 = 7$
Hence, the coordinates are (2,7)
Now, equation of tangent paasing through (2,7) and with slope m = 2 is
y = mx + c
7 = 2 X 2 + c
c = 7 - 4 = 3
So,
y = 2 X x+ 3
y = 2x + 3
So, the equation of tangent is y - 2x = 3

Question:15(b) Find the equation of the tangent line to the curve $y = x^2 -2x +7$ which is perpendicular to the line $5y - 15x = 13.$

Perpendicular to line $5y - 15x = 13.\Rightarrow y = 3x + \frac{13}{5}$ means $slope \ of \ tangent = \frac{-1}{slope \ of \ line}$
We know that the equation of the line is
y = mx + c
on comparing with the given equation we get the slope of line m = 3 and c = 13/5
$slope \ of \ tangent = \frac{-1}{slope \ of \ line} = \frac{-1}{3}$
Now, we know that the slope of the tangent at a given point to given curve is given by $\frac{dy}{dx}$
Given the equation of curve is
$y = x^2 - 2x +7$
$\frac{dy}{dx} = 2x - 2 = \frac{-1}{3}\\ \\ x = \frac{5}{6}$
Now, when $x = \frac{5}{6}$ , $y = (\frac{5}{6})^2 - 2(\frac{5}{6}) +7 = \frac{25}{36} - \frac{10}{6} + 7 = \frac{217}{36}$
Hence, the coordinates are $(\frac{5}{6} ,\frac{217}{36})$
Now, the equation of tangent passing through (2,7) and with slope $m = \frac{-1}{3}$ is
$y = mx+ c\\ \frac{217}{36}= \frac{-1}{3}\times \frac{5}{6} + c\\ c = \frac{227}{36}$
So,
$y = \frac{-1}{3}x+\frac{227}{36}\\ 36y + 12x = 227$
Hence, equation of tangent is 36y + 12x = 227

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

Slope of tangent = $\frac{dy}{dx} = 21x^2$
When x = 2
$\frac{dy}{dx} = 21x^2 = 21(2)^{2} = 21 \times4 = 84$
When x = -2
$\frac{dy}{dx} = 21x^2 = 21(-2)^{2} = 21 \times4 = 84$
Slope is equal when x= 2 and x = - 2
Hence, we can say that both the tangents to curve $y = 7x^3 + 11$ is parallel

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.

Given equation of curve is $y = x ^3$
Slope of tangent = $\frac{dy}{dx} = 3x^2$
it is given that the slope of the tangent is equal to the y-coordinate of the point
$3x^2 = y$
We have $y = x ^3$
$3x^2 = x^3\\ 3x^2 - x^3=0\\ x^2(3-x)=0\\ x= 0 \ \ \ \ \ \ \ \ and \ \ \ \ \ \ \ \ \ \ x = 3$
So, when x = 0 , y = 0
and when x = 3 , $y = x^3 = 3^3 = 27$

Hence, the coordinates are (3,27) and (0,0)

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

Tangent passes through origin so, (x,y) = (0,0)
Given equtaion of curve is $y = 4x ^ 3 - 2x ^5$
Slope of tangent =

$\frac{dy}{dx} = 12x^2 - 10x^4$
Now, equation of tangent is
$Y-y= m(X-x)$
at (0,0) Y = 0 and X = 0
$-y= (12x^3-10x^4)(-x)$
$y= 12x^3-10x^5$
and we have $y = 4x ^ 3 - 2x ^5$
$4x^3-2x^5= 12x^3-10x^5$
$8x^5 - 8x^3=0\\ 8x^3(x^2-1)=0\\ x=0\ \ \ \ \ \ and \ \ \ \ \ \ \ x = \pm1$
Now, when x = 0,

$y = 4(0) ^ 3 - 2(0) ^5 = 0$
when x = 1 ,

$y = 4(1) ^ 3 - 2(1) ^5 = 4-2=2$
when x= -1 ,

$y = 4(-1) ^ 3 - 2(-1) ^5 = -4-(-2)=-4+2=-2$
Hence, the coordinates are (0,0) , (1,2) and (-1,-2)

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

parellel to x-axis means slope is 0
Given equation of curve is
$x^2 + y^2 - 2x - 3 = 0$
Slope of tangent =
$-2y\frac{dy}{dx} = 2x -2\\ \frac{dy}{dx} = \frac{1-x}{y} = 0\\ x= 1$
When x = 1 ,

$-y^2 = x^2 -2x-3= (1)^2-2(1)-3 = 1-5=-4$
$y = \pm 2$
Hence, the coordinates are (1,2) and (1,-2)

Question:20 Find the equation of the normal at the point $( am^2 , am^3 )$ for the curve $ay ^2 = x ^3.$

Given equation of curve is
$ay ^2 = x ^3\Rightarrow y^2 = \frac{x^3}{a}$
Slope of tangent

$2y\frac{dy}{dx} = \frac{3x^2 }{a} \Rightarrow \frac{dy}{dx} = \frac{3x^2}{2ya}$
at point $( am^2 , am^3 )$
$\frac{dy}{dx} = \frac{3(am^2)^2}{2(am^3)a} = \frac{3a^2m^4}{2a^2m^3} = \frac{3m}{2}$
Now, we know that
$Slope \ of \ normal = \frac{-1}{Slope \ of \ tangent} = \frac{-2}{3m}$
equation of normal at point $( am^2 , am^3 )$ and with slope $\frac{-2}{3m}$
$y-y_1=m(x-x_1)\\ y-am^3 = \frac{-2}{3m}(x-am^2)\\ 3ym - 3am^4 = -2(x-am^2)\\ 3ym +2x= 3am^4+2am^2$
Hence, the equation of normal is $3ym +2x= 3am^4+2am^2$

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

Equation of given curve is
$y = x^3 + 2x + 6$
Parellel to line $x + 14y + 4 = 0 \Rightarrow y = \frac{-x}{14} -\frac{4}{14}$ means slope of normal and line is equal
We know that, equation of line
y= mx + c
on comparing it with our given equation. we get,
$m = \frac{-1}{14}$
Slope of tangent = $\frac{dy}{dx} = 3x^2+2$
We know that
$Slope \ of \ normal = \frac{-1}{Slope \ of \ tangent} = \frac{-1}{3x^2+2}$
$\frac{-1}{3x^2+2} = \frac{-1}{14}$
$3x^2+2 = 14\\ 3x^2 = 12 \\ x^2 = 4\\ x = \pm 2$
Now, when x = 2, $y = (2)^3 + 2(2) + 6 = 8+4+6 =18$
and
When x = -2 , $y = (-2)^3 + 2(-2) + 6 = -8-4+6 =-6$
Hence, the coordinates are (2,18) and (-2,-6)
Now, the equation of at point (2,18) with slope $\frac{-1}{14}$
$y-y_1=m(x-x_1)\\ y-18=\frac{-1}{14}(x-2)\\ 14y - 252 = -x + 2\\ x+14y = 254$
Similarly, the equation of at point (-2,-6) with slope $\frac{-1}{14}$

$y-y_1=m(x-x_1)\\ y-(-6)=\frac{-1}{14}(x-(-2))\\ 14y + 84 = -x - 2\\ x+14y + 86= 0$
Hence, the equation of the normals to the curve $y = x^3 + 2x + 6$ which are parallel
to the line $x + 14y + 4 = 0.$

are x +14y - 254 = 0 and x + 14y +86 = 0

Question:22 Find the equations of the tangent and normal to the parabola $y ^2 = 4 ax$ at the point $(at ^2, 2at).$

Equation of the given curve is
$y ^2 = 4 ax$

Slope of tangent = $2y\frac{dy}{dx} = 4a \Rightarrow \frac{dy}{dx} = \frac{4a}{2y}$
at point $(at ^2, 2at).$
$\frac{dy}{dx}= \frac{4a}{2(2at)} = \frac{4a}{4at} = \frac{1}{t}$
Now, the equation of tangent with point $(at ^2, 2at).$ and slope $\frac{1}{t}$ is
$y-y_1=m(x-x_1)\\ y-2at=\frac{1}{t}(x-at^2)\\ yt - 2at^2 = x - at^2\\ x-yt +at^2 = 0$

We know that
$Slope \ of \ normal = \frac{-1}{Slope \ of \ tangent} = -t$
Now, the equation of at point $(at ^2, 2at).$ with slope -t
$y-y_1=m(x-x_1)\\ y-2at=(-t)(x-at^2)\\ y - 2at = -xt + at^3\\ xt+y -2at -at^3 = 0$
Hence, the equations of the tangent and normal to the parabola

$y ^2 = 4 ax$ at the point $(at ^2, 2at).$ are
$x-yt+at^2=0\ \ \ \ and \ \ \ \ xt+y -2at -at^3 = 0 \ \ respectively$

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

Let suppose, Curve $x = y^2$ and xy = k cut at the right angle
then the slope of their tangent also cut at the right angle
means,
$\left ( \frac{dy}{dx} \right )_a \times \left ( \frac{dy}{dx} \right )_b = -1$ -(i)
$2y\left ( \frac{dy}{dx} \right )_a = 1 \Rightarrow \left ( \frac{dy}{dx} \right )_a = \frac{1}{2y}$
$\left ( \frac{dy}{dx} \right )_b = \frac{-k}{x^2}$
Now these values in equation (i)
$\frac{1}{2y} \times \frac{-k}{x^2} = -1\\ -k = -2yx^2\\ k =2(xy)(x)\\ k = 2k(k^{\frac{2}{3}}) \ \ \ \ \left ( x = y^2 \Rightarrow y^2y = k \Rightarrow y = k^{\frac{1}{3}} \ and \ x = k^{\frac{2}{3}} \right ) \\ 2(k^{\frac{2}{3}}) = 1\\ \left ( 2(k^{\frac{2}{3}}) \right )^3 = 1^3\\ 8k^2 = 1$
Hence proved

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 $(x_0 , y_0 )$

Given equation is
$\frac{x^2 }{a^2} - \frac{y^2 }{b^2 }= 1 \Rightarrow y^2a^2 = x^2b^2 -a^2b^2$
Now ,we know that
slope of tangent = $2ya^2\frac{dy}{dx} = 2xb^2 \Rightarrow \frac{dy}{dx} = \frac{xb^2}{ya^2}$
at point $(x_0 , y_0 )$
$\frac{dy}{dx} = \frac{x_0b^2}{y_0a^2}$
equation of tangent at point $(x_0 , y_0 )$ with slope $\frac{xb^2}{ya^2}$
$y-y_1=m(x-x_1)\\ y-y_0=\frac{x_0b^2}{y_0a^2}(x-x_0)\\ yy_0a^2-y_0^2a^2 = xx_0b^2-x_0^2b^2\\ xx_0b^2 - yy_0a^2 = x_0^2b^2-y_0^2a^2$
Now, divide both sides by $a^2b^2$
$\frac{xx_0}{a^2} - \frac{yy_0}{b^2} = \left ( \frac{x_0^2}{a^2} - \frac{y_0^2}{b^2} \right )$
$=1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left ( \because \frac{x_0^2}{a^2}-\frac{y_0^2}{b^2 } = 1\right )$
$\frac{xx_0}{a^2} - \frac{yy_0}{b^2} = 1$
Hence, the equation of tangent is

$\frac{xx_0}{a^2} - \frac{yy_0}{b^2} = 1$
We know that
$Slope \ of \ normal= \frac{-1}{slope \ of \ tangent } = -\frac{y_0a^2}{x_0b^2}$
equation of normal at the point $(x_0 , y_0 )$ with slope $-\frac{y_0a^2}{x_0b^2}$
$y-y_1=m(x-x_1)\\ y-y_0=-\frac{y_0a^2}{x_0b^2}(x-x_0)\\ \frac{y-y_0}{y_0a^2} + \frac{x-x_0}{x_0b^2} = 0$

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

Parellel to line $4x - 2y + 5 = 0 \Rightarrow y = 2x + \frac{5}{2}$ means the slope of tangent and slope of line is equal
We know that the equation of line is
y = mx + c
on comparing with the given equation we get the slope of line m = 2 and c = 5/2
Now, we know that the slope of the tangent at a given point to given curve is given by $\frac{dy}{dx}$
Given the equation of curve is
$y = \sqrt{3x-2}$
$\frac{dy}{dx} = \frac{1}{2}.\frac{3}{\sqrt{3x-2}}=\frac{3}{2\sqrt{3x-2}}$
$\frac{3}{2\sqrt{3x-2}} = 2\\ 3^2 = (4\sqrt{3x-2})^2\\ 9 = 16(3x-2)\\ 3x-2=\frac{9}{16}\\ 3x = \frac{9}{16} +2\\ 3x= \frac{41}{16}\\ x = \frac{41}{48}$
Now, when

$x = \frac{41}{48}$ , $y = \sqrt{3x-2} \Rightarrow y = \sqrt{3\times\frac{41}{48}-2 } = \sqrt{\frac{41}{16}-2}=\sqrt\frac{9}{16 } = \pm \frac{3}{4}$

but y cannot be -ve so we take only positive value
Hence, the coordinates are

$\left ( \frac{41}{48},\frac{3}{4} \right )$
Now, equation of tangent paasing through

$\left ( \frac{41}{48},\frac{3}{4} \right )$ and with slope m = 2 is
$y - y_1=m(x-x_1)\\ y-\frac{3}{4}=2(x-\frac{41}{48})\\ 48y-36=2(48x-41)\\ 48x-24y=41-18\\ 48x-24y=23$
Hence, equation of tangent paasing through $\left ( \frac{41}{48},\frac{3}{4} \right )$ and with slope m = 2 is 48x - 24y = 23

Question:26 The slope of the normal to the curve $y = 2x ^2 + 3 \sin x \: \: at \: \: x = 0$ is
(A) 3 (B) 1/3 (C) –3 (D) -1/3

Equation of the given curve is
$y = 2x ^2 + 3 \sin x$
Slope of tangent = $\frac{dy}{dx} = 4x +3 \cos x$
at x = 0
$\frac{dy}{dx} = 4(0) +3 \cos 0= 0 + 3$
$\frac{dy}{dx}= 3$
Now, we know that
$Slope \ of \ normal = \frac{-1}{\ Slope \ of \ tangent} = \frac{-1}{3}$
Hence, (D) is the correct option

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) (C) (1, – 2) (D) (– 1, 2)

The slope of the given line $y = x+1$ is 1
given curve equation is
$y^2 = 4 x$
If the line is tangent to the given curve than the slope of the tangent is equal to the slope of the curve
The slope of tangent = $2y\frac{dy}{dx} = 4 \Rightarrow \frac{dy}{dx} = \frac{2}{y}$
$\frac{dy}{dx} = \frac{2}{y} = 1\\ y = 2$
Now, when y = 2, $x = \frac{y^2}{4} = \frac{2^2}{4} = \frac{4}{4} = 1$
Hence, the coordinates are (1,2)

Hence, (A) is the correct answer

## More About NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3

As the questions in the Class 12 Maths chapter 6 exercise 6.3 deal with an application of derivatives, it is better for the students to revise the basic derivatives of trigonometric functions, exponential functions and some other special functions and rules and properties related to the derivatives. Out of the 27 problems in the Class 12th Maths chapter, 6 exercise 6. 3 question 26 is to find the slope of the normal to a given curve and question 27 asks to find the points for which a line is a tangent to the given curve.

Also Read| Application of Derivatives Class 12 Notes

## Benefits of NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3

• Solving exercise 6.3 Class 12 Maths will be beneficial for both the CBSE board exam and JEE Main exam.

• One question from NCERT solutions for Class 12 Maths chapter 6 exercise 6.3 can be expected for the CBSE Class 12 Maths board paper.

• The applications of tangents will be used in Class 12 Physics and Chemistry also.

Also see-

• NCERT Exemplar Solutions Class 12 Maths Chapter 6

• NCERT Solutions for Class 12 Maths Chapter 6

## NCERT Solutions Subject Wise

• NCERT Solutions Class 12 Chemistry

• NCERT Solutions for Class 12 Physics

• NCERT Solutions for Class 12 Biology

• NCERT Solutions for Class 12 Mathematics

## Subject Wise NCERT Exemplar Solutions

• NCERT Exemplar Class 12 Maths

• NCERT Exemplar Class 12 Physics

• NCERT Exemplar Class 12 Chemistry

• NCERT Exemplar Class 12 Biology