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Class 12 NCERT Solutions- Mathematics Part I – Chapter 6 Application of Derivatives -Exercise 6.3 | Set 2
  • Last Updated : 06 Apr, 2021

Chapter 6 Application of Derivatives -Exercise 6.3 | Set 1

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

(i) y = x4– 6x3 + 13x2 – 10x+ 5 at (0, 5)

(ii) y = x4 – 6x3 + 13x2 – 10x + 5 at (1, 3)

(iii) y = x3 at(1, 1)

(iv) y = x2 at(0, 0)

(v) x = cos t, y = sin t at t = π/4



Solution:

(i) Given curve

y = x4 – 6x3 + 13x2 – 10x + 5

Given point, (0, 5)

dy/dx = 4x3 – 18x2 + 26x – 10

dy/dx = 4(0)3 – 18(0)2 + 26(0) – 10

dy/dx = -10,

-dx/dy = 1/10 

Now, with the help of points slope form 

y – y1 = m(x – x1)

y – 5 = -10(x – 0)

y + 10x = 5 is the required equation of the tangent 

For equation of normal,

y – y1 = \frac{-dx}{dy}(x-x_1)

y – 5 = \frac{1}{10}(x-0)

10y – x – 50 is the equation of normal.

(ii) Given curve: y = x4 – 6x3 + 13x2 – 10x + 5

Given point is (1, 3)



dy/dx = 4x3 – 18x2 + 26x – 10 

dy/dx = 4(1)3 – 18(1)2 + 26(1) – 10

dy/dx = 4 – 18 + 26 – 10 = 2

dy/dx = 2       

-dx/dy = -1/2  

Using point slope form, equation of tangent is 

y – y1 = dy/dx(x – x1)

y – 3 = 2(x – 1)

y – 2x = 1 is the equation of tangent.

Using point slope form, equation of normal is

y –  y1 = -dx/dy(x – 1)

y – 3 = -1/2(x – 1)

2y – 6 = -x + 1

2y + x = 7 is the equation of normal.

(iii) Given curve : y = x3

Given point is (1, 1)

dy/dx = 3x2

dy/dx = 3(1)2 = 3

dy/dx = 3 & -dx/dy = -1/3

Using point slope form, equation of tangent is y – y1 = dy/dx(x – x1)

y – y1 = dy/dx(x – x1)

y – 1 = 3(x – 1)

y – 3x + 2 = 0 is the equation of tangent

Using point slope form, equation of normal is 

y – y1 = \frac{-dx}{dy}(x-x_1)

y – 1 = \frac{-1}{3}(x-1)

3y – 3 = -x + 1

3y + x = 4 is the equation of normal.

(iv) Given curve: y = x2

Given point (0, 0)

dy/dx = 2x

dy/dx = 0

dy/dx = 0 & -dx/dy = not defined is 

y – y1 = dy/dx(x – x1)

y – 0 = 0(x – 0)

y = 0 is the equation of tangent.

Using point slope form, equation of normal is 

y – y1 \frac{-dx}{dy}(x-x_1)              -(1)

-dx\dy is undefined, so we can write eq(1) as 

\frac{-dy}{dx}(y-y_1)=x-x_1

Now putting dy/dx = 0 we get

0(y – 0) = x-0

x = 0 is the equation of normal.

(v) Equation of curve: x = cos t and y = sin t

Point t = π/4

\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}          -(1)

\frac{dx}{dt}=\frac{d(\cos t)}{dt}=-\sin t

\frac{dy}{dt}=\frac{d(\sin t)}{dt}=\cos t

On putting these values in eq(1), we get

\frac{dy}{dx}=\frac{\cos t}{-\sin t}=-\cot t

\frac{dy}{dx}=-\cot \frac{π}{4}=-1

dy/dx = -1 & -dx/dy = 1  

Now for t = π/4,

y1 = sin t = sin(π/4) = 1/√2

x1 = cos t = cos(π/4) = 1/√2

The point is (1/√2, 1/√2) 

y – y1 \frac{dy}{dx}(x-x_1)

y – (1/√2) = -1(x – 1/√2)

y – 1/√2 = -x + 1/√2

x + y = √2 is the equation of normal is 

y-y_1=\frac{-dx}{dy}(x-x_1)

y – 1/√2 = 1(x – 1/√2)

x = y is the equation of normal.

Question 15. Find the equation of the tangent line to the curve y = x2 – 2x + 7 which is 

(i) Parallel to line 2x – y + 9 = 0 

(ii) Perpendicular to the line 5y – 15x = 13

Solution:

Given curve: y = x2 – 2x + 7

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

dy/dx = 2x – 2         -(1)

(i) Tangent is parallel to 2x – y + 9 = 0 that means,

Slope of tangent = slope of 2x – y + 9 = 0

y = 2x + 9

Slope = 2            -(Comparing with y = mx + e)

dy/dx = slope = 2

2x – 2 = 2

x1 = 2

Corresponding to x1 = 2,

y1 = x12 – 2x1 + 7 

y1 = (2)2 – 2(2) + 7

y1 = 7

The point of contact is (2, 7).

Using point slope form, equation of tangent is 

y – y1 \frac{dy}{dx}(x-x_1)

y – 7 = 2(x – 2)

y – 2x = 3 is the equation of tangent. 

(ii) Tangent is perpendicular to the line 5y – 15x = 13

That means (slope of tangent) x (slope of line) = -1

For, slope of line 5y – 15x = 13

5y = 15x + 13

y = 3x + 13/15

Slope = 3

(Slope of tangent) x 3 = -1

Slope of tangent =-1/3 

Now, y = x2 – 2x + 7

dy/dx = 2x – 2

On comparing dy/dx with slope, we get

2x – 2 = -1/3

6x – 6 = -1

6x = 5

x1 = 5/6

For x1 = 5/6,          

y1 = x12 – 2x1 + 7 

y1 = (5/6)12 – 2(5/6)1 + 7 

y1 = 217/36 

Now using point slope form, equation of tangent is 

y – y1 = m(x – x1)

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

\frac{36y-217}{36}=-1\frac{(6x-5)}{6.3}

36y – 217 = -12x + 10

36y + 12x = 227 is the required equation of tangent.

Question 16. Show that the tangent to the curve y = 7x3 + 11 at the points where x = 2 and x = -2 are parallel.

Solution:

Given curve: y = 7x3 + 11

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

dy/dx = 21x2

dy/dx = 21(2)2 = 84

The slopes at x – 2 & -2 are the same,

Hence the tangent will be parallel to each other.

Question 17. Find the points on the curve y = x3 at which the slope of the tangent is equal to the y-coordinates of the point.

Solution:

Given curve: y = x3

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

dy/dx = 3x2          -(1)

Now let us assume that the point is (x1, y1)

dy/dx = 3x12 

Also, slope of tangent at (x1, y1) is equal to y1.

So, 3x12 = y1         -(2)

Also, (x1, y1) lies on y = x3 x3,so

y1 = x13           -(3)

From eq(2) & (3)

3x12 = x13  

3x12 = x13 = 0

x12(3 – x1) = 0 

For x1 = 0, y1 = x13 = (0)3 = 0

One such point is (0, 0)

For x1 = 3, y1. = (3)3 = 27

Second point is (3, 27)

Question 18. For the curve y = 4x3 – 2x5, find all the points at which the tangent passes through the origin.

Solution:

Given curve: y = 4x3 – 3x5

Clearly at x = 0, y = 0, i.e the curve passes through origin. 

Now the tangent also passes through origin. 

Equation of a line passing through origin is y = mx.

Now tangent is touching the curve, so 

y = mx will satisfy in curve.

mx = 4x3 – 2x5         -(1)

Now dy/dx = 12x2 – 10x4

Also m is the slope of tangent, so

m = 12x2 – 10x4          -(2)

From eq(1) & (2),

(12x2 – 10x4)x = 4x3 – 3x5

x3(12 – 10x2) = x3(4 – 2x2)          -(3)

For the first point, x = 0

For x1 = 0, y1 = 4x13 – 2x15 = 0 

So, (0, 0) is one such point

Now for other roots of 3

12 – 10x2 = 4 – 2x2

8 = 8x2

x2 = 1

x = ±1

For x2 = 1, y2 = 4x23 – 2x25 = 4(1)3 – 2(1)5 = 2 

For x3 = -1,  y3 = 4x33 – 2x35 = 4(-1)3 – 2(-1)5 = -2   

The other points are(1, 2) & (-1, -2)

Question 19. Find the points on the curve x2 + y2 – 2x – 3 = 0 at which the tangents are parallel to the x-axis.

Solution:

Given curve: x2 + y2 – 2x – 3 = 0

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

3x + 2y(dy/dx) – 2 – 0 = 0

x+y\frac{dy}{dx}=1 \frac{dy}{dx}=\frac{1-x}{y}

Given that the tangent are parallel to x-axis,

So, dy/dx = slope = 0

1 – x/y = 0

For x1 = 1,

x12 + y12 – 2x1 – 3 = 0      

(1)2 + y12 – 2(1) – 3 = 0      

 y12 = 4     

y12 = ≠2

The points are (1, 2) & (1, -2)

Question 20. Find the equation of the normal at the point (am2, am3) for the curve ay2 = x3.

Solution:

Given curve: ay2 = x3

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

2ay.dy/dx = 3x2

\frac{dy}{dx}=\frac{3x^2}{2ay}  & \frac{-dx}{dy}=\frac{-2ay}{3x^2}

\frac{-dx}{dy}=\frac{-2a.am^3}{3.(am^2)^2}=\frac{-2}{3m}

So, by point slope form, equation of normal is,

y-y_1=\frac{-dx}{dy}(x-x_1)    

y-am^3=\frac{-2}{3m}(x-am^2)

3my – 3am4 = -2x + 2am2

Hence, 3my + 2x = 2am2 + 3am4 is the required equation of normal.

Question 21. Find the equation of the normals to the curve y = x3 + 2x + 6 which are parallel to the line x + 14y + 4 = 0.

Solution:

Given curve: y = x3 + 2x + 6

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

\frac{dy}{dx}=3x^2+2

\frac{-dx}{dy}=\frac{-1}{3x^2+2}          -(Slope of normal)

Now, the normal are parallel to x + 14y + 4 = 0

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

13 = 3x2 + 2            

3x2 = 12 ⇒  x2 = 4

x = ±2

x1 = 2 & x2 = -2

For x1 = 2; y1 x_1^3+2x_1+6  = (2)3 + 2(2) + 6 = 18

For x2 = -2; y2 x_2^2+2x_2+6  = (-2)3 + 2(-2) + 6 = -6

Normal through (2,18) is 

y – 18 = \frac{-1}{14}(x-2)

14y – 252 = -x + 2

14y + x = 254 is one such equation.

Normal through (-2, -6) is 

y + 6 = \frac{-1}{14}(x+2)

14y + 84 = -x – 2

14y + x + 86 = 0 is the other equation of normal.

Question 22. Find the equations of tangent and normal to the parabola y2 = 4ax at the point (at2, 2at).

Solution:

Given parabola: y2 = 4ax

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

2y.\frac{dy}{dx}=4a

\frac{dy}{dx}=\frac{2a}{y}  ;\frac{-dx}{dy}=\frac{-y}{2a}

\frac{dy}{dx}=\frac{2a}{2at}=\frac{1}{t}

Now, by point slope form, equation of tangent is,

y – y1 \frac{dy}{dx}(x-x_1)

y-2at=\frac{1}{t}(x-at^2)

ty = x + at2 is the equation of tangent to the parabola y2 = 4ax at (at2, 2at)

Now \frac{-dx}{dy}=\frac{-2at}{2a}=-t

Now, by point slope form, equation of normal is,

y – y1 \frac{-dx}{dy}(x-x_1)

y – 2at = -t(x – at2)

y + xt = 2at + at3 is the equation of normal to the parabola y2 = 4ax at (at2, 2at)

Question 23. Prove that the curves x = y2 and xy = k cut at right angles if 8k2 = 1.

Solution:

Given curves: x = y2 & xy = k

Two curves intersect at right angles f the tangents through their point 

intersection is perpendicular to each other.

Now if tangents are perpendicular their product of their slopes will be equal to -1.

Curve 1: x = y2

1 = 2y.dy/dx

dy/dx = 1/2 y = m1          -(1)

Curve 2: xy = k

y = k/x

\frac{-k}{x^2}=m_2

Let’s find their point of intersection 

x = y2 & xy = k

k/y = y2 

y = k1/3

x = y2      

x = k2/3

The point is (k2/3, k1/3)

m1 = 1/2k1/3

For curves to be intersecting each other at right angles,

m1m2 = -1

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

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

8k2 = 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 (xo, yo).

Solution:

Given curve: \frac{x^2}{a^2}-\frac{y^2}{b^2}=1

On differentiating both sides with respect to x,

\frac{2x}{a^2}-\frac{2y}{b^2}.\frac{dy}{dx}=0    

\frac{dy}{dx}=\frac{xb^2}{ya^2}\&\frac{-dx}{dy}=\frac{-ya^2}{xb^2}

Now, \frac{dy}{dx}=\frac{x_ob^2}{y_0a^2}=m_T

Equation of tangent by point slope form is,

y-y_1=m_T(x-x_1)

y-y_0=\frac{x_0b^2}{y_0a^2}(x-x_0)

\frac{yy_0-y_0^2}{b^2}=\frac{xx_0-x_0^2}{a^2}

\frac{yy_0}{b^2}-\frac{xx_0}{a^2}=\frac{y_0^2}{a^2}=-1         -((x0,y0) lie on \frac{x^2}{a^2}-\frac{b^2}{b^2}=1 )

\frac{xx_0}{a^2}-\frac{yy_0}{b^2}=1  is the equation of tangent.

Now, \frac{-dx}{dy}=\frac{-y_0a^2}{x_0b^2}=m_N

Equation of normal by point slope form is,

y-y_1=m_N(x-x_1)

y-y_0=\frac{-y_0a^2}{x_0b^2}(x-x_0)

x0b2y – x0b2y0 = -y0a2x + y0a2x0

x0b2y + y0a2x = x0y0(a2 + b2) is the equation of normal

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.

Solution:

Given curve: y=\sqrt{3x-2}

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

\frac{dy}{dx}=\frac{1}{2\sqrt{3x-2}}.3

Now, it is given that the tangent is parallel to the line 4x – 2y + 5 = 0, 

so their slopes must be equal.

Slope of 2y = 4x + 5 is 2. 

So,

\frac{dy}{dx}=\frac{3}{2\sqrt{3x-2}}=2

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

9/16 = 3x – 2

x1 = 41/48

Now, y_1=\sqrt{3x_1-2}

y_1=\sqrt{3.\frac{41}{48}-2}=\frac{3}{4}

The point is (41/48, 3/4)

Now by point slope form, the equation of tangent will be 

y – y1 = m(x – x1)

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

y-\frac{3}{4}=2x-\frac{41}{24}

24y – 48x + 23 = 0 is the required equation of tangent.

Question 26. The slope of the normal to the curve y = 2x2 + 3 sin x at x = 0 is 

(A) 3          (B) 1/3          (C)-3          (D) -1/3

Solution:

Given curve: y = 2x2 + 3 sin x

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

dy/dx = 4x + 3cos x

\frac{-dx}{dy}=\frac{-1}{4x+3\cos x}                  -(Slope of normal)

\frac{-dx}{dy}=\frac{-1}{4(0)+3\cos 0}=\frac{-1}{3}

Hence, the correct option is D.

Question 27. The line y = x + 1 is a tangent to the curve y2 = 4x at the point

(A) (1, 2)          (B) (2, 1)          (C) (1, -2)          (D) (-1, 2)

Solution:

Given curve: y2 = 4x

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

2y\frac{dy}{dx}=4        

\frac{dy}{dx}=\frac{2}{y}

y = x + 1 is tangent, slope is 1, so, 2/y = 2

y1 = 2,

y12 = 4x1 

x1\frac{2^2}{4}    

x1 = 1

So, (1, 2) is the point.

Hence, the correct option is A.

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