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

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

Edited By Ramraj Saini | Updated on Dec 03, 2023 08:43 PM IST | #CBSE Class 12th

NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.3

NCERT Solutions for Exercise 6.3 Class 12 Maths Chapter 6 Application of Derivatives are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. 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.

12th class Maths exercise 6.3 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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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

Answer:

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

Answer:

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.

Answer:

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.

Answer:

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

Answer:

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

Answer:

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.

Answer:

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).

Answer:

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

Answer:

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

Answer:

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

Answer:

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 }

Answer:

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

Answer:

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

Answer:

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)

Answer:

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)

Answer:

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)

Answer:

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)

Answer:

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

Answer:

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

Answer:

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.

Answer:

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 .

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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).

Answer:

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.

Answer:

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 )

Answer:

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 .

Answer:

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

Answer:

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)

Answer:

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.

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Key Features Of NCERT Solutions for Exercise 6.3 Class 12 Maths Chapter 6

  • Comprehensive Coverage: The solutions encompass all the topics covered in ex 6.3 class 12, ensuring a thorough understanding of the concepts.
  • Step-by-Step Solutions: In this class 12 maths ex 6.3, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
  • Accuracy and Clarity: Solutions for class 12 ex 6.3 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
  • Conceptual Clarity: In this 12th class maths exercise 6.3 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
  • Inclusive Approach: Solutions for ex 6.3 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
  • Relevance to Curriculum: The solutions for class 12 maths ex 6.3 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

Also see-

NCERT Solutions Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Question (FAQs)

1. What is the equation of the tangent to a curve y=f(x) at (x0,y0)?

y-y0=f’(x0)(x-x0) 

2. Give the equation of normal to the tangent y-y0=f’(x0)(x-x0)

(y-y0)f’(x0)+(x-x0)=0

3. Equation of a tangent with slope=0 st (x0, y0) is

The slope =0. Therefore the equation of tangent is y=y0

4. The slope of a tangent is infinity, then what is its equation at (x0,y0)?

The equation of tangent with infinite slope is x=x0

5. Normal is ……………...to the tangent

Normal is perpendicular to the tangent

6. What is the relation between the slope of a tangent and the normal to the tangent?

The slope of normal is the negative of the inverse of the slope of the tangent.

7. Give the number of questions discussed in Exercise 6.3 Class 12 Maths

27 questions are answered in the NCERT Solutions for Class 12 Maths chapter 6 exercise 6.3. For more questions refer to NCERT exemplar. Following NCERT syllabus will be useful for CBSE board exams.

8. How many solved examples are given in the topic tangents and normals?

There are 7 solved examples explained before the NCERT book Class 12 Maths chapter 6 exercise 6.3.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?
i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

You can use them people also used problem

Read Complete Answer
Divyan 06 August,2024
i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

The Medhavi National Scholarship Program, under the Human Resources & Development Mission (HRDM), offers financial assistance to meritorious students through a scholarship exam. To be eligible, candidates must be between 16 and 40 years old as of the last date of registration and have at least passed the 10th grade from a recognized board. Higher qualifications, such as 11th/12th grade, graduation, post-graduation, or a diploma, are also acceptable.

To apply, download the Medhavi App from the Google Play Store, sign up, and read the detailed notification about the scholarship exam. Complete the registration within the app, take the exam from home using the app, and receive your results within two days. Following this, upload the necessary documents and bank account details for verification. Upon successful verification, the scholarship amount will be directly transferred to your bank account.

The scholarships are categorized based on the marks obtained in the exam: Type A for those scoring 60% or above, Type B for scores between 50% and 60%, and Type C for scores between 40% and 50%. The cash scholarships range from Rs. 2,000 to Rs. 18,000 per month, depending on the exam and the marks obtained.

Since you already have a 12th-grade qualification with 84%, you meet the eligibility criteria and can apply for the Medhavi Scholarship exam. Preparing well for the exam can increase your chances of receiving a higher scholarship.

Read Complete Answer
Yuvan 01 September,2024
i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Read Complete Answer
Ashvadha 12 July,2024
Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Read Complete Answer
kumardurgesh1802 10 July,2024
I have completed my 12th with arts cbse. I want to repeat my 12th with pcmb. How to become a private candidate and what are deadline to be able to appear in 2025 exam?

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

  • No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
  • You have to appear for the 2025 12th board exams.
  • Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
  • Aim to register before late October to avoid extra fees.
  • Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

Good luck with your studies!

Read Complete Answer
kumardurgesh1802 10 July,2024
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Popular CBSE Class 12th Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

 Option 2)

0.16\; J
Option 3)

1.00\; J

 Option 4)

0.67\; J
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A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

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2.45×10−3 kg

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 6.45×10−3 kg
Option 3)

 9.89×10−3 kg

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12.89×10−3 kg

 
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2,000 \; J - 5,000\; J

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200 \, \, J - 500 \, \, J
Option 3)

2\times 10^{5}J-3\times 10^{5}J

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20,000 \, \, J - 50,000 \, \, J
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A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

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K/2\,

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\; K\;
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In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

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 Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced
Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

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67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .
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How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

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0.02

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3.125 × 10-2
Option 3)

1.25 × 10-2

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2.5 × 10-2
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If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

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increase two fold
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remain unchanged

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be a function of the molecular mass of the substance.
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With increase of temperature, which of these changes?

Option 1)

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Option 3)

Fraction of solute present in water

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Mole fraction.
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Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

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6.023 × 1022
Option 3)

half that in 8 g He

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558.5 × 6.023 × 1023
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A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

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more than 3 but less than 6
Option 3)

more than 6 but less than 9

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more than 9
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