Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

Class 12 RD Sharma Solutions – Chapter 16 Tangents and Normals – Exercise 16.2 | Set 2

  • Last Updated : 14 Jul, 2021

Question 7. Find the equation of the normal to the curve ay2 = x3 at the point (am2, am3).

Solution:

We have,

ay2 = x3

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

2aydy/dx = 3x2

dy/dx = 3x2/2ay

Slope of tangent = \left(\frac{dy}{dx} \right)_{\left( a m^2 , a m^3 \right)} =\frac{3 a^2 m^4}{2 a^2 m^3}=\frac{3m}{2}

Given (x1, y1) = (am2, am3)

The equation of normal is,

y – y1 = -1/m (x – x1)

y – a m3 = -2m/3 (x – am2)

3my – 3am4 = – 2x + 2am2

2x + 3my – am2 (2 + 3m2) = 0

Question 8. The equation of the tangent at (2, 3) on the curve y2 = ax3 + b is y = 4x − 5. Find the values of a and b.

Solution:

We have,

y2 = ax3 + b

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

2y dy/dx = 3ax2

dy/dx = 3ax2/2y

Slope of tangent, m = \left( \frac{dy}{dx} \right)_{\left( 2, 3 \right)} =\frac{12a}{6}=2a

The equation of tangent is given by y – y1 = m (tangent) (x – x1)

Now compare the slope of a tangent with the given equation

2a = 4

a = 2

Now (2, 3) lies on the curve, these points must satisfy

32 = 2 (23) + b

b = – 7

Question 9. Find the equation of the tangent line to the curve y = x2 + 4x − 16 which is parallel to the line 3x − y + 1 = 0.

Solution:

We have,

y = x2 + 4x − 16

Let (a, b) be the point of intersection of both the curve and the tangent.

Since (a, b) lies on curve, we get

b = a2 + 4a − 16

Now, x2 + 4x − 16

dy/dx = 2x + 4

Slope of tangent = \left( \frac{dy}{dx} \right)_{\left( a , b \right)}=2 a +4

Given that the tangent is parallel to the line we have,

Slope of tangent = Slope of the given line

=> 2a + 4 = 3

=> 2 a = -1

=> a = -1/2

From eq(1), we get

b = 1/4 – 2 – 16 = -71/4

Now, slope of tangent, m = 3

(a, b) = (-1/2, -71/4)

The equation of tangent is,

y – y1 = m (x – x1)

y + 71/4 = 3 (x + 1/2)

\frac{4y + 71}{4} = 3\left( \frac{2x + 1}{2} \right)

4y + 71 = 12x + 6

12x – 4y – 65 = 0

Question 10. Find an equation of normal line to the curve y = x3 + 2x + 6 which is parallel to the line x+ 14y + 4 = 0.

Solution:

We have,

y = x3 + 2x + 6

Let (a, b) be a point on the curve where we need to find the normal.

Slope of the given line = -1/14

Since the point lies on the curve, we get

b = a3 + 2a + 6

Now, y = x3 + 2x + 6

dy/dx = 3 x2 + 2

Slope of the tangent, m = \left( \frac{dy}{dx} \right)_{\left( a , b \right)} =3 {a}^2 +2

Slope of the normal = \frac{- 1}{3 {a}^2 + 2}

Given that, slope of the normal = slope of the given line, we have

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

3a2 + 2 = 14

3a2 = 12

a2 = 4

a = ±2

So, b = 18 or -6.

And slope of the normal = -1/14

When a = 2 and b = 18, we have

y – y1 = m (x – x1)

y – 18 = -1/14 (x – 2)

14y – 252 = -x + 2

x + 14y – 254 = 0

When a = -2 and b = -6, we have

y – y1 = m (x – x1)

y + 6 = -1/14 (x + 2)

14y + 84 = -x – 2

x + 14y + 86 = 0

Question 11. Determine the equation(s) of tangent (s) line to the curve y = 4x3 − 3x + 5 which are perpendicular to the line 9y + x + 3 = 0.

Solution:

Let (a, b) be a point on the curve where we need to find the tangent(s).

Slope of the given line = -1/9

Since, tangent is perpendicular to the given line,

Slope of the tangent = \frac{- 1}{\left( \frac{- 1}{9} \right)}  = 9

Hence, b = 4 a3 – 3 a + 5

Now, y = 4 x3 – 3x + 5

dy/dx = 12 x2 – 3

Slope of the tangent = \left( \frac{dy}{dx} \right)_{\left( a , b \right)} =12 {a}^2 - 3

Given that, slope of the tangent = slope of the perpendicular line

12a2 – 3 = 9

12a2 = 12

a2 = 1

a = ±1

So, b = 6 or 4.

Thus, slope of tangent = 9.

When a = 1 and b = 6, we have

y – y1 = m (x – x1)

y – 6 = 9 (x – 1)

y – 6 = 9x – 9

9x – y – 3 = 0

When a = -1 and b = 4, we have

y – y1 = m (x – x1)

y – 4 = 9 (x + 1)

y – 4 = 9x + 9

9x – y + 13 = 0

Question 12. Find the equation of a normal to the curve y = x loge x which is parallel to the line 2x − 2y + 3 = 0.

Solution:

Slope of the given line is 1.

Let (a, b) be the point where the tangent is drawn to the curve.

Hence, b = a loge a  . . . . (1)

Now, y = x loge x

dy/dx = x × 1/x + loge x(1) = 1 + loge x1

Slope of tangent = 1 + log a

Slope of normal = \frac{- 1}{1 + \log_e a}

Given that, slope of normal = slope of the given line.

\frac{- 1}{1 + \log_e a} = 1

=> -1 = 1 + log a

=> – 2 = log a

=> a = e-2 

From (1), we have

Now, b = e-2 (-2) = -2 e-2

Given, (x1, y1) = (e-2, -2 e-2)

The equation of normal is,

y + 2/e2 = 1(x – 1/e2)

y + 2/e2 = x – 1/e2

x – y = 3/e2

Question 13. Find the equation of the tangent line to the curve y = x2 − 2x + 7 

(i) which is parallel to the line 2x − y + 9 = 0?

Solution:

We have, y = x2 − 2x + 7 

On differentiating both sides, we get

dy/dx = 2x – 2

The equation of the line is 2x – y + 9 = 0

So the slope of line is 2.

According to the question,

=> 2x – 2 = 2

=> 2x = 4

=> x = 2

=> y = 22 − 2(2) + 7 = 4 – 4 + 7 = 7

As (x1, y1) is (2, 7), the equation of tangent is,

y – 7 = 2(x – 2)

y – 7 = 2x – 4

y – 2x – 3 = 0

(ii) which is perpendicular to the line 5y − 15x = 13.

Solution:

We have, y = x2 − 2x + 7

On differentiating both sides, we get

dy/dx = 2x – 2

The equation of the line is 5y − 15x = 13.

=> y = 3x + 13/5

So the slope of line is 3.

According to the question,

=> 2x – 2= -1/3

=> 6x – 6 = -1

=> x = 5/6

And y = 217/36.

As (x1, y1) is (5/6, 217/36), the equation of tangent is,

y – y1 = m (x – x1)

y – 217/36 = (-1/3) (x – 5/6)

36y -217 = -12x + 10

36y + 12x – 227 = 0

Question 14. Find the equations of all lines having slope 2 and that are tangent to the curve y = 1/x – 3, x ≠ 3.

Solution:

Let (a , b) be the point where the tangent is drawn to this curve.

Since, the point lies on the curve, hence b = 1/(a – 3)

\frac{dy}{dx} = \frac{- 1}{\left( x - 3 \right)^2}

Slope of tangent, m = \left( \frac{dy}{dx} \right)=\frac{- 1}{\left( a - 3 \right)^2}

Slope of the tangent} = 2

\frac{- 1}{\left( a - 3 \right)^2} = 2

(a – 3)2 = – 2

a – 3 = √-2, which does not exist because 2 is negative.

So, there does not exist any such tangent.

Question 15. Find the equations of all lines of slope zero and that are tangent to the curve y = \frac{1}{x^2 - 2x + 3} .

Solution:

Slope of the given tangent is 0.

Let (a, b) be a point where the tangent is drawn to the curve.

Since, the point lies on the curve, hence b = \frac{1}{{a}^2 - 2 a + 3}  . . . . (1)

\frac{dy}{dx} = \frac{\left( x^2 - 2x + 3 \right)\left( 0 \right) - \left( 2x - 2 \right)1}{\left( x^2 - 2x + 3 \right)^2} = \frac{- 2x + 2}{\left( x^2 - 2x + 3 \right)^2}

Slope of tangent = \frac{- 2 a + 2}{\left( {a}^2 - 2 a + 3 \right)^2}

Given that, slope of tangent = slope of the given line, 

\frac{- 2 a + 2}{\left( {a}^2 - 2 a + 3 \right)^2} = 0

=> -2 a + 2 = 0

=> 2a = 2

=> a = 1

From (1), we get

Now, b = \frac{1}{1 - 2 + 3} = \frac{1}{2}

(a, b) = (1, 1/2) 

The equation of tangent is,

y – y1 = m (x – x1)

y – 1/2 = 0 (x – 1)

y = 1/2

Question 16. Find the equation of the tangent to the curve y = \sqrt{3x - 2}  which is parallel to the 4x − 2y + 5 = 0.

Solution:

We have,

y = \sqrt{3x - 2}

Let (a, b) be the point where the tangent is drawn to the curve y = \sqrt{3x - 2}

On differentiating both sides, we get

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

Slope of tangent at (a, b) = \frac{3}{2\sqrt{3 a - 2}}

Slope of line 4x − 2y + 5 = 0 is 2.

Given that, slope of tangent = slope of the given line

\frac{3}{2\sqrt{3 a - 2}} = 2

3 = 4\sqrt{3 a - 2}

9 = 16 (3a – 2)

9/16  = 3a – 2

3a = 9/16 + 2 

a = 41/48

Now, b = \sqrt{\frac{123}{48} - 2} = \sqrt{\frac{27}{48}} = \sqrt{\frac{9}{16}} = \frac{3}{4}

Therefore, (a, b) = (41/48, 3/4)

The equation of tangent is,

y – y1 = m (x – x1)

y – 3/4 = 2 (x – 41/48)

(4y – 3)/4 = 2 (48x – 41)/48

24y – 18 = 48x – 41

48x – 24y – 23 = 0

Question 17. Find the equation of the tangent to the curve x2 + 3y − 3 = 0, which is parallel to the line y= 4x − 5.

Solution:

Suppose (a, b) be the required point.

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

So, slope of the line = 4

Since (a, b) lies on the curve, we get a2 + 3b − 3 = 0  . . . . (1)

Now,

2x + 3dy/dx = 0

dy/dx = -2x/3

Slope of tangent, m= \left( \frac{dy}{dx} \right)_{\left( a , b \right)} =\frac{- 2a}{3}

Given that tangent is parallel to the line, So we get,

Slope of tangent, m = slope of the given line

=> -2a/3 = 4 

 => a = -6

From (1), we get

=> 36 + 3b – 3 = 0

=> 3b = – 33

=> b = – 11

(a, b) = (-6, -11)

The equation of tangent is,

y – y1 = m (x – x1)

y + 11 = 4 (x + 6)

y + 11 = 4x + 24

4x – y + 13 = 0

Question 18. Prove that \left( \frac{x}{a} \right)^n + \left( \frac{y}{b} \right)^n = 2  touches the straight line \frac{x}{a} + \frac{y}{b} = 2  for all n ∈ N, at the point (a, b) ?

Solution:

We have,

\left( \frac{x}{a} \right)^n + \left( \frac{y}{b} \right)^n = 2

On differentiating both sides, we get

\frac{n}{a} \left( \frac{x}{a} \right)^{n - 1} + \frac{n}{b} \left( \frac{y}{b} \right)^{n - 1} \frac{dy}{dx} = 0

\frac{n}{b} \left( \frac{y}{b} \right)^{n - 1} \frac{dy}{dx} = \frac{- n}{a} \left( \frac{x}{a} \right)^{n - 1}

\frac{dy}{dx} = \frac{- n}{a} \left( \frac{x}{a} \right)^{n - 1} \times \frac{b}{n} \left( \frac{b}{y} \right)^{n - 1} = \frac{- b}{a} \left( \frac{bx}{ay} \right)^{n - 1}

Slope of tangent = \left( \frac{dy}{dx} \right)_{\left( a, b \right)} =\frac{- b}{a} \left( \frac{b * a}{a * b} \right)^{n - 1} =\frac{- b}{a}

The equation of tangent is,

y – b = -b/a (x – a)

ya – ab = – xb + ab

xb + ya = 2ab

\frac{x}{a} + \frac{y}{b} = 2

So, the given line touches the given curve at the given point.

Hence proved.

Question 19. Find the equation of the tangent to the curve x = sin 3t, y = cos 2t at t = π/4.

Solution:

We have,

x = sin 3t, y = cos 2t

dx/dt = 3 cos3t and dy/dt = -2sin2t

\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{- 2 \sin 2t}{3 \cos 3t}

Slope of tangent, m= \left( \frac{dy}{dx} \right)_{t = \frac{\pi}{4}} =-\frac{- 2 \sin \left( \frac{\pi}{2} \right)}{3 \cos \left( \frac{3\pi}{4} \right)}=\frac{- 2}{\frac{- 3}{\sqrt{2}}}=\frac{2\sqrt{2}}{3}

x1 = sin 3π/4 = 1/√2 and y1 = cos π/2 = 0

So, (x1, y1) = (1/√2, 0)

The equation of tangent is,

y – y1 = m (x – x1)

y - 0 = \frac{2\sqrt{2}}{3}\left( x - \frac{1}{\sqrt{2}} \right)

3y = 2√2 x – 2

2√2 x – 3y – 2 = 0

Question 20. At what points will be tangents to the curve y = 2x3 − 15x2 + 36x − 21 be parallel to x-axis? Also, find the equations of the tangents to the curve at these points?

Solution:

We have,

y = 2x3 − 15x2 + 36x − 21

The slope of x – axis is 0.

Let (a, b) be the required point.

Since (a, b) lies on the curve, we get

b = 2a3 − 15a2 + 36a − 21   . . . . (1)

Also, we have

dy/dx = 6 x2 – 30x + 36

Slope of tangent at (a, b) = \left( \frac{dy}{dx} \right)_{\left( a , b \right)} = 6 {a}^2 - 30 a + 36

Given that the slope of the tangent = slope of the x-axis, we have

=> 6a2 – 30a + 36 = 0

=> a2 – 5a + 6 = 0

=> (a – 2) (a – 3) = 0

=> a = 2 or a= 3

=> b = 7 or 6

When a = 2 and b = 7, the equation is,

y – y1 = m (x – x1)

y – 7 = 0 (x – 2)

y = 7

When a = 3 and b = 6, the equation is,

y – y1 = m (x – x1)

y – 6 = 0 (x – 3)

y = 6

Question 21. Find the equation of the tangents to the curve 3x2 – y2 = 8, which passes through the point (4/3, 0).

Solution:

We have,

3x2 – y2 = 8  . . . . (1)

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

6x – 2y dy/dx = 0

2y dy/dx = 6x

dy/dx = 6x/2y

dy/dx = 3x/y

Let tangent at (h, k) pass through (4/3, 0). Since, (h, k) lies on (1), we get

3 h2 – k2 = 8 . . . (ii)

Slope of tangent at (h, k) = 3h/k

The equation of tangent at (h, k) is given by,

y – k = 3h/k (x – h) 

Also,

=> 0 – k = (3h/k) (4/3 – h)

=> -k = 4h/k – 3h2/k

=> – k2 = 4h – 3h2

=> 8 – 3 h2 = 4h – 3 h2

=> 8 = 4h

=> h = 2

Also we get,

=> 12 – k2 = 8

=> k2 = 4

=> k = ±2

So, the points on curve (i) at which tangents pass through (4/3, 0) are (2, ±2).

When h = 2 and k = 2, the equation is,

y – 2 = (6/2) (x – 2)

3x – y – 4 = 0

When h = 2 and k = –2, the equation is,

y + 2 = (6/-2) (x – 2)

3x + y – 4 = 0


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!