Class 12 RD Sharma Solutions – Chapter 16 Tangents and Normals – Exercise 16.2 | Set 1

Question 1. Find the equation of the tangent to the curve √x + √y = a at the point (a²/4, a²/4).

Solution:

We have,

√x + √y = a

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

dy/dx = -√y/√x

Given, (x₁, y₁) = (a²/4, a²/4), 

Slope of tangent, m = 

The equation of tangent is,

y – y₁ = m (x – x₁)

y – a²/4 = –1(x – a²/4)

y – a²/4 = –x + a²/4

x + y = a²/2

Question 2. Find the equation of the normal to y = 2x³ − x² + 3 at (1, 4).

Solution:

We have,

y = 2x³ − x² + 3

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

dy/dx = 6x² – 2x

Slope of tangent =  = 6 (1)² – 2 (1) = 4

Slope of normal = – 1/Slope of tangent = – 1/4

Given, (x₁, y₁) = (1, 4), 

The equation of normal is,

y – y₁ = m (x – x₁)

y – 4 = -1/4 (x – 1)

4y – 16 = – x + 1

x + 4y = 17

Question 3. Find the equation of the tangent and the normal to the following curve at the indicated point:

(i) y = x⁴ − bx³ + 13x² − 10x + 5 at (0, 5)

Solution:

We have,

y = x⁴ − bx³ + 13x² − 10x + 5

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

dy/dx = 4x³ – 3bx² + 26x – 10

Slope of tangent, m=  = -10

Given, (x₁, y₁) = (0, 5) 

The equation of tangent is,

y – y₁ = m (x – x₁)

y – 5 = – 10 (x – 0)

y – 5 = -10x

y + 10x – 5 = 0

The equation of normal is,

y – y₁ = -1/m (x – x₁)

y – 5 = 1/10 (x – 0)

10y – 50 = x

x – 10y + 50 = 0

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

Solution:

We have,

y = x⁴ − 6x³ + 13x² − 10x + 5

When x = 1, we have y = 1 – 6 + 13 – 10 + 5 = 3

So, (x₁, y₁) = (1, 3) 

Now, y = x⁴ − 6x³ + 13x² − 10x + 5

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

dy/dx = 4 x³ – 18 x² + 26x – 10

Slope of tangent, m =  = 4 – 18 + 26 – 10 = 2

The equation of tangent is,

y – y₁ = 2 (x – x₁)

y – 3 = 2 (x – 1)

y – 3 = 2x – 2

2x – y + 1 = 0

The equation of normal is,

y – y₁ = -1/m (x – x₁)

y – 3 = -1/2 (x – 1)

2y – 6 = – x + 1

x + 2y – 7 = 0

(iii) y = x² at (0, 0)

Solution:

We have,

y = x²

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

dy/dx = 2x

Given, (x₁, y₁) = (0, 0) 

Slope of tangent, m=  = 2 (0) = 0

The equation of tangent is, 

y – y₁ = m (x – x₁)

y – 0 = 0 (x – 0)

y = 0

The equation of normal is,

y – y₁ = -1/m (x – x₁)

y – 0 = -1/0 (x – 0)

x = 0

(iv) y = 2x² − 3x − 1 at (1, −2)

Solution:

We have,

y = 2x² − 3x − 1

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

dy/dx = 4x – 3

Given, (x₁, y₁) = (1, -2) 

Slope of tangent, m =  = 4 – 3 = 1

The equation of tangent is,

y – y₁ = m (x – x₁)

y + 2 = 1 (x – 1)

y + 2 = x – 1

x – y – 3 = 0

The equation of normal is,

y – y₁ = -1/m (x – x₁)

y + 2 = -1 (x – 1)

y + 2 = -x + 1

x + y + 1 = 0

(v)  at (2, -2)

Solution:

We have,

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

= 

= 

Given, (x₁, y₁) = (2, -2)

Slope of tangent, m=  =  = -2

The equation of tangent is,

y – y₁ = m (x – x₁)

y + 2 = -2 (x – 2)

y + 2 = -2x + 4

2x + y – 2 = 0

The equation of normal is,

y – y₁ = -1/m (x – x₁)

y + 2 = 1/2 (x – 2)

2y + 4 = x – 2

x – 2y – 6 = 0

(vi) y = x² + 4x + 1 at x = 3 

Solution:

We have,

y = x² + 4x + 1

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

dy/dx = 2x + 4

When x = 3, y = 9 + 12 + 1 = 22

So, (x₁, y₁) = (3, 22) 

Slope of tangent, m=  = 10

The equation of tangent is,

y – y₁ = m (x – x₁)

y – 22 = 10 (x – 3)

y – 22 = 10x – 30

10x – y – 8 = 0

The equation of normal is,

y – y₁ = -1/m (x – x₁)

y – 22 = -1/10 (x – 3)

10y – 220 = – x + 3

x + 10y – 223 = 0

(vii)  at (a cos θ, b sin θ)

Solution:

We have,

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

dy/dx = -x b²/y a²

Slope of tangent, m= 

Given, (x₁, y₁) = (a cos θ, b sin θ)

The equation of tangent is,

y – y₁ = m (x – x₁)

y – b sin θ = -bcosθ/asinθ  (x – a cos θ)

ay sin θ – ab sin² θ = -bx cos θ + ab cos² θ

bx cos θ + ay sin θ = ab

On dividing by ab, we get

x/a cosθ + y/b sinθ = 1

The equation of normal is,

y – y₁ = -1/m (x – x₁)

y – b sin θ = asinθ/bcosθ  (x – a cos θ)

by cos θ – b² sin θ cos θ = ax sin θ – a² sin θ cos θ

ax sin θ – by cos θ = (a² – b²) sin θ cos θ

On dividing by sin θ cos θ, we get

ax sec θ – by cosec θ = a² – b²

(viii)  at (a sec θ, b tan θ)

Solution:

We have,

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

dy/dx = x b²/y a²

Slope of tangent}, m= 

Given, (x₁, y₁) = (a sec θ, b tan θ)

The equation of tangent is,

y – y₁ = m (x – x₁)

y – b tan θ =  (x – a sec θ)

ay sin θ – ab(sin² θ/cos θ) = bx – (ab/cos θ) 

ay sin θ cos θ – ab sin² θ = bx cos θ – ab

bx cos θ – ay sin θ cos θ = ab (1 – sin² θ)

bx cos θ – ay sin θ cos θ = ab cos² θ

On dividing by ab cos² θ, we get

x/a sec θ – y/b tan θ = 1

The equation of normal is,

y – y₁ = -1/m (x – x₁)

y – b tan θ = -a sin θ/b (x – a sec θ)

by – b² tan θ = -ax sin θ + a² tan θ

ax sin θ + by = (a² + b²) tan θ

On dividing by tan θ, we get

ax cos θ + by cot θ = a² + b²

(ix) y² = 4ax at (a/m², 2a/m)

Solution:

We have,

y² = 4ax

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

dy/dx = 2a/y

Given, (x₁, y₁) = (a/m², 2a/m)

Slope of tangent =  = m

The equation of tangent is, 

y – y₁ = m (x – x₁)

my – 2a = m² x – a

m² x – my + a = 0

The equation of normal is,

y – y₁ = -1/m (x – x₁)

m³ y – 2a m² = – m² x + a

m² x + m³ y – 2a m² – a = 0

(x) c² (x² + y²) = x²y² at (c/cos θ, c/sin θ)

Solution:

We have,

c² (x² + y²) = x²y²

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

2x c² + 2y c²(dy/dx) = x² 2y(dy/dx) + 2x y²

dy/dx(2y c² – 2 x² y) = 2x y² – 2x c²

Slope of tangent, m= 

= 

= 

= 

= -cos³ θ/ sin³ θ 

Given, (x₁, y₁) = (c/cos θ, c/sin θ)

The equation of tangent is,

y – y₁ = m (x – x₁)

sin² θ (y sin θ – c) = -cos² θ (x cos θ – c)

y sin³ θ – c sin² θ = – x cos³ θ + c cos² θ

x cos³ θ + y sin³ θ = c ( sin² θ + cos² θ)

x cos³ θ + y sin³ θ = c

The equation of normal is,

y – y₁ = -1/m (x – x₁)

sin³ θ – ycos³ θ = 2c[-cos (2θ)/sin(2θ)]

sin³ θ – y cos³ θ = -2c cot 2θ

sin³ θ – y cos³ θ + 2c cot 2θ = 0

(xi) xy = c² at (ct, c/t)

Solution:

We have,

xy = c²

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

dy/dx = – y/x

Given, (x₁, y₁) = (ct, c/t)

Slope of tangent, m= 

The equation of tangent is,

y – y₁ = m (x – x₁)

yt² – ct = -x + ct

x + y t² = 2ct

The equation of normal is,

y – y₁ = -1/m (x – x₁)

y – c/t – t²(x – ct)

yt – c = t³ x – c t⁴

x t³ – yt = c t⁴ – c

(xii) at (x₁, y₁)

Solution:

We have,

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

dy/dx = – x b²/y a²

Slope of tangent, m= 

The equation of tangent is,

y – y₁ = m (x – x₁)

y – y₁ = – x₁ b²/y₁ a²(x – x₁)

y y₁ a² – y₁² a² = -x x₁ b² + x₁² b²

x x₁ b² + y y₁ a² = x₁² b² + y₁² a₂  . . . . (1)

Given (x₁, y₁) lies on the curve, we get

x₁² b² + y₁² a² = a² b²

Substituting this in (1), we get

x x₁ b² + y y₁ a² = a² b²

On dividing this by a² b², we get

The equation of normal is,

y – y₁ = m (x – x₁)

y – y₁ = y₁ a²/x₁ b²(x – x₁)

y x₁ b² – x₁ y₁ b² = x y₁ a² – x₁ y₁ a²

x y₁ a² – y x₁ b² = x₁ y₁ a² – x₁ y₁ b²

x y₁ a² – y x₁ b² = x₁ y₁ (a² – b²)

On dividing by x₁ y₁, we get

(xiii) at (x₀ , y₀)

Solution:

We have,

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

dy/dx = x b²/y a²

Slope of tangent, m= 

The equation of tangent is,

y – y₁ = m (x – x₁)

y – y₀ = x₀ b²/y₀ a²(x – x₀)

y y₀ a² – y₀² a² = x x₀ b² – x₀² b²

x x₀ b² – y y₀ a² = x₀² b² – y₀² a²  . . . . (1)

x₀²b²– y₀²a² = a² b² 

Substituting this in eq(1), we get,

x x₀ b² – y y₀ a² = a² b²

Dividing this by a² b², we get

The equation of normal is,

y – y₁ = m (x – x₁)

y – y₀ = y₀ a²/x₀ b²(x – x₀)

y x₀ b² – x₀ y₀ b² = -x y₀ a² + x₀ y₀ a²

x y₀ a² + y x₀ b² = x₀ y₀ a² + x₀ y₀ b²

x y₀ a² + y x₀ b² = x₀ y₀ (a² + b²)

Dividing by x₀ y₀, we get

(xiv) at (1, 1)

Solution:

We have,

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

Slope of tangent, m= = -1

Given, (x₁, y₁) = (1, 1)

The equation of tangent is,

y – y₁ = m (x – x₁)

y – 1 = -1 (x – 1)

y – 1 = -x + 1

x + y – 2 = 0

The equation of normal is,

y – y₁ = -1/m (x – x₁)

y – 1 = 1 (x – 1)

y – 1 = x – 1

y – x = 0

(xv) x² = 4y at (2, 1)

Solution:

We have,

x² = 4y 

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

2x = 4dy/dx

dy/dx = x/2

Slope of tangent, m=  = 2/2 = 1

Given, (x₁, y₁) = (2, 1)

The equation of tangent is,

y – y₁ = m (x – x₁)

y – 1 = 1 (x – 2)

y – 1 = x – 2

x – y – 1 = 0

The equation of normal is,

y – y₁ = -1/m (x – x₁)

y – 1 = – 1 (x – 2)

y – 1 = – x + 2

x + y – 3 = 0

(xvi) y² = 4x at (1, 2)

Solution:

We have,

y² = 4x

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

2y (dy/dx) = 4

dy/dx = 2/y

Slope of tangent, m= = 2/2 = 1

Given, (x₁, y₁) = (1, 2)

The equation of tangent is,

y – y₁ = m (x – x₁)

y – 2 = 1 (x – 1)

y – 2 = x – 1

x – y + 1 = 0

The equation of normal is,

y – y₁ = -1/m (x – x₁)

y – 2 = -1 (x – 1)

y – 2 = -x + 1

x + y – 3 = 0

(xvii) 4x² + 9y² = 36 at (3 cos θ, 2 sin θ)

Solution:

We have,

4x² + 9y² = 36

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

8x + 18y dy/dx = 0

18y dy/dx = – 8x

dy/dx = -8x/18y = -4x/9y

Slope of tangent, m = 

The equation of tangent is,

y – y₁ = m (x – x₁)

y – 2 sin θ = -2 cos θ/3 sin θ(x – 3 cos θ)

3y sin θ – 6 sin² θ = -2x cos θ + 6 cos² θ

2x cos θ + 3y sin θ = 6 (cos² θ + sin² θ)

2x cos θ + 3y sin θ = 6

The equation of normal is,

y – y₁ = -1/m (x – x₁)

y – 2 sin θ = -3 sin θ/2 cos θ(x – 3 cos θ)

2y cos θ – 4 sin θ cos θ = 3x sin θ – 9 sin θ cos θ

3x sin θ – 2y cos θ – 5sin θ cos θ = 0

(xviii) y² = 4ax at (x₁, y₁)

Solution:

We have,

y² = 4ax

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

2y dy/dx = 4a

dy/dx = 2a/y

At (x₁, y₁), we have

Slope of tangent = = m

The equation of tangent is,

y – y₁ = m (x – x₁)

y y₁ – y₁² = 2ax – 2a x₁

y y₁ – 4a x₁ = 2ax – 2a x₁

y y₁ = 2ax + 2a x₁

y y₁ = 2a (x + x₁)

The equation of normal is,

y – y₁ = -1/m (x – x₁)

y – y₁ = -y₁/2a (x – x₁)

(xix) at (√2a, b)

Solution:

We have,

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

dy/dx = x b²/y a²

Slope of tangent, m= 

The equation of tangent is,

y – y₁ = m (x – x₁)

y – b = √2b/a(x – √2a)

ay – ab = √2 bx – 2ab

√2 bx – ay = ab

The equation of normal is, 

y – y₁ = -1/m (x – x₁)

y – b = – a/√2b(x – √2a)

√2 by – √2 b² = – ax + √2 a² 

ax + √2 by = √2 b² + √2 a²

ax/√2 + by = a² + b²

Question 4. Find the equation of the tangent to the curve x = θ + sin θ, y = 1 + cos θ at θ = π/4.

Solution:

We have,

x = θ + sin θ, y = 1 + cos θ

 and 

Slope of tangent, m = 

= 

= 

= 

= 

= 1 – √2 

Given, (x₁, y₁) = (π/4 + sin π/4, 1 + cos π/4) = (π/4 + 1/√2, 1 + 1/√2)

The equation of tangent is,

y – y₁ = m (x – x₁)

y – (1 + 1/√2) = (1 – √2) [x – (π/4 + 1/√2)]

y – 1 – 1/√2 = (1 – √2) (x – π/4 – 1/√2)

Question 5. Find the equation of the tangent and the normal to the following curve at the indicated points. 

(i) x = θ + sin θ, y = 1 + cos θ at θ = π/2

Solution:

We have,

x = θ + sin θ and y = 1 + cos θ

dx/dθ = 1 + cos θ and dy/dθ = -sinθ

= 

Slope of tangent, m= 

= -1/(1 + 0)

= -1

Given, (x₁, y₁) = (π/2 + sin π/2, 1 + cos π/2) = (π/2 + 1, 1)

The equation of tangent is,

y – y₁ = m (x – x₁)

y – 1 = -1 (x – π/2 – 1)

2y – 2 = – 2x + π + 2

x + 2y – π – 4 = 0

The equation of normal is,

y – y₁ = -1/m (x – x₁)

y – 1 = 1 (x – π/2 -1)

2y – 2 = 2x – π – 2

2x – 2y = π

(ii)  at t = 1/2

Solution:

We have,

dx/dt = 

= 

dy/dt = 

 = 

Slope of tangent, m= 

Given, (x₁, y₁) = 

The equation of tangent is,

y – y₁ = m (x – x₁)

80y – 16a = 65x – 26a

65x – 80y – 10a = 0

13x – 16y – 2a = 0

The equation of normal is,

y – y₁ = -1/m (x – x₁)

65y – 13a = – 80x + 32a

80x + 65y – 45a = 0

16x + 13y – 9a = 0

(iii) x = at², y = 2at at t = 1

Solution:

We have,

x = at², y = 2at

dx/dt = 2at and dy/dt = 2a

Slope of tangent, m=  = 1

Given, (x₁ , y₁) = (a, 2a)

The equation of tangent is,

y – y₁ = m (x – x₁)

y – 2a = 1 (x – a)

y – 2a = x – a

x – y + a = 0

Equation of normal:

y – y₁ = -1/m (x – x₁)

y – 2a = – 1 (x – a)

y – 2a = – x + a

x + y = 3a

(iv) x = a sec t, y = b tan t at t 

Solution:

We have,

x = a sec t, y = b tan t

dx/dt = a sect tant and dy/dt = b sec²t

Slope of tangent, m = 

Given (x₁, y₁) = (a sec t, b tan t)

The equation of tangent is,

y – y₁ = m (x – x₁)

y – b tan t = (b/a) cosec t (x – a sec t)

ay sin t cos t – ab sin² t = bx cos t – ab

bx cos t – ay sin t cos t – ab (1 – sin² t) = 0

bx cos t – ay sin t cos t = ab cos² t

On dividing by cos² t, we get

bx sec t – ay tan t = ab

The equation of normal is,

y – y₁ = -1/m (x – x₁)

y – b tant = -a/b sint(x – asect)

ycost − bsint = − a/b​sint(xcost − a)

by cos t – b² sin t = – ax sin t cos t + a² sin t

ax sin t cos t + by cos t = (a² + b²) sin t

On dividing both sides by sin t, we get

ax cos t + by cot t = a² + b²

(v) x = a(θ + sin θ), y = a(1 − cos θ) at θ

Solution:

We have,

x = a(θ + sin θ), y = a(1 − cos θ)

dx/dθ = a(1 + cosθ) and dy/dθ = asinθ

= 

= 

= 

= tan θ/2   . . . . (1)

Slope of tangent, m= 

Given (x₁, y₁) = [a(θ + sin θ), a(1 − cos θ)]

The equation of tangent is,

y – y₁ = m (x – x₁)

y – a (1 – cos θ) = tan θ/2 [x – a (θ + sin θ)]

y − a(2sin²θ/2​) = xtanθ/2​ − aθtanθ/2 ​− atanθ/2​sinθ

y − 2asin²θ/2 ​ =(x − aθ)tan θ/2 − 2asin²θ/2​

y = (x – aθ) tan θ/2

The equation of normal is,

y – a (1 – cos θ) = -cot θ/2 [x – a (θ + sin θ)]

tanθ/2​[y − a(2sin²θ/2​)] = −x + aθ + asinθ

tanθ/2​[y − a{2(1 − cos²θ/2​)}] = −x + aθ + asinθ

tan θ/2 (y – 2a) + a (2sin θ/2 cosθ/2 = -x + aθ + a sinθ

tan θ/2 (y – 2a) + a sin θ = -x + aθ + a sin θ

tan θ/2 (y – 2a) = – x + aθ

tan θ/2 (y – 2a) + x – θ = 0

(vi) x = 3 cos θ − cos³ θ, y = 3 sin θ − sin³ θ

Solution:

We have,

x = 3 cos θ − cos³ θ, y = 3 sin θ − sin³ θ

dx/dθ = -3sin θ + 3 cos²θ sin θ and dy/dθ = 3 cos θ – 3 sin²θ cos θ

= 

= 

=  cos³ θ/ -sin³ θ 

= tan³ θ

So the equation of the tangent at θ is,

y – 3 sin θ + sin³ θ = -tan³ θ (x – 3 cos θ + cos³ θ)

4 (y cos³ θ – x sin³ θ) = 3 sin 4θ

So the equation of normal at θ is,

y – 3 sin θ + sin³ θ= (1/tan³ θ) (x – 3 cos θ + cos³ θ)

sin³ θ – x cos³ θ = 3sin⁴ θ – sin⁶ θ – 3cos⁴ θ + cos⁶ θ

Question 6. Find the equation of the normal to the curve x² + 2y² − 4x − 6y + 8 = 0 at the point whose abscissa is 2?

Solution:

Given that abscissa = 2. i.e., x = 2

x² + 2y² − 4x − 6y + 8 = 0  . . . . (1)

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

2x + 4y dy/dx – 4 – 6 dy/dx = 0

dy/dx(4y – 6) = 4 – 2x

When x = 2, we get

4 + 2y² – 8 – 6y + 8 = 0

2y² – 6y + 4 = 0

y² – 3y + 2 = 0

y = 2 or y = 1

m (tangent) at x = 2 is 0

Normal is perpendicular to tangent so, m₁m₂ = –1

m (normal) at x = 2 is 1/0, which is undefined.

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

x = 2