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

Chapter 16 of RD Sharma's Class 12 mathematics textbook focuses on Tangents and Normals, with Exercise 16.2 Set 2 specifically dealing with advanced problems related to these concepts.

This section builds upon the fundamental principles of calculus and analytical geometry, challenging students to apply their knowledge to solve complex problems involving tangent lines and normal lines to curves.

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 log_e 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 log_e a  . . . . (1)

Now, y = x log_e x

dy/dx = x × 1/x + log_e x(1) = 1 + log_e 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

Summary

Exercise 16.2 Set 2 of Chapter 16 in RD Sharma's Class 12 mathematics textbook covers advanced problems on tangents and normals. The questions in this set require students to apply differentiation techniques, solve equations, and interpret geometric relationships. Key concepts include finding equations of tangents and normals, determining points of intersection, analyzing parallel and perpendicular lines, and working with various curve types including polynomial, exponential, logarithmic, and conic sections.