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

  • Last Updated : 14 Jul, 2021

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

\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}}\frac{dy}{dx} = 0

dy/dx = -√y/√x

Given, (x1, y1) = (a2/4, a2/4), 

Slope of tangent, m = (\frac{dy}{dx})_{\left(\frac{a^2}{4},\frac{a^2}{4}\right)}=\frac{-\sqrt{\frac{a^2}{4}}}{\sqrt{\frac{a^2}{4}}}=-1

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 = \left( \frac{dy}{dx} \right)_{\left( 1, 4 \right)}  = 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= \left( \frac{dy}{dx} \right)_{\left( 0, 5 \right)}  = -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 = \left( \frac{dy}{dx} \right)_{\left( 1, 3 \right)}   = 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= \left(\frac{dy}{dx} \right)_{\left( 0, 0 \right)}   = 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 = \left( \frac{dy}{dx} \right)_{\left( 1, - 2 \right)}  = 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) y^2 = \frac{x^3}{4 - x}   at (2, -2)

Solution:

We have,

y^2 = \frac{x^3}{4 - x}

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

2y \frac{dy}{dx} = \frac{\left( 4 - x \right)\left( 3 x^2 \right) - x^3 \left( - 1 \right)}{\left( 4 - x \right)^2}

\frac{12 x^2 - 3 x^3 + x^3}{\left( 4 - x \right)^2}

\frac{12 x^2 - 2 x^3}{\left( 4 - x \right)^2}

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

Slope of tangent, m= \left( \frac{dy}{dx} \right)_{\left( 2, - 2 \right)}   = \frac{48 - 16}{- 16}      = -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= \left( \frac{dy}{dx} \right)_{x = 3}   = 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) \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1      at (a cos θ, b sin θ)

Solution:

We have,

\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

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

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

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

dy/dx = -x b2/y a2

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

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) \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1      at (a sec θ, b tan θ)

Solution:

We have,

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

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

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

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

dy/dx = x b2/y a2

Slope of tangent}, m= \left(\frac{dy}{dx} \right)_{\left( a \sec \theta, b \tan \theta \right)} =\frac{a \sec \theta \left( b^2 \right)}{b \tan \theta \left( a^2 \right)}=\frac{b}{a \sin \theta}

Given, (x1, y1) = (a sec θ, b tan θ)

The equation of tangent is,

y – y1 = m (x – x1)

y – b tan θ = \frac{b}{a \sin \theta}      (x – a sec θ)

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

\frac{ay \sin \theta \cos \theta - ab \sin^2 \theta}{\cos \theta} = \frac{bx \cos \theta - ab}{\cos \theta}

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

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

dy/dx = 2a/y

Given, (x1, y1) = (a/m2, 2a/m)

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

The equation of tangent is, 

y – y1 = m (x – x1)

y - \frac{2a}{m} = m \left( x - \frac{a}{m^2} \right)

\frac{my - 2a}{m} = m\left( \frac{m^2 x - a}{m^2} \right)

my – 2a = m2 x – a

m2 x – my + a = 0

The equation of normal is,

y – y1 = -1/m (x – x1)

y - \frac{2a}{m} = \frac{- 1}{m}\left( x - \frac{a}{m^2} \right)

\frac{my - 2a}{m} = \frac{- 1}{m}\left( \frac{m^2 x - a}{m^2} \right)

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

\frac{dy}{dx} = \frac{x y^2 - x c^2}{y c^2 - x^2 y}

Slope of tangent, m= \left( \frac{dy}{dx} \right)_{\left( \frac{c}{\cos \theta}, \frac{c}{\sin \theta} \right)}

\frac{\frac{c^3}{\cos \theta \sin^2 \theta} - \frac{c^3}{\cos \theta}}{\frac{c^3}{\sin\theta} - \frac{c^3}{\cos^2 \theta \sin\theta}}

\frac{\frac{1 - \sin^2 \theta}{\cos\theta \sin^2 \theta}}{\frac{\cos^2 \theta - 1}{\cos^2 \theta \sin\theta}}

\frac{co s^2 \theta}{\cos \theta \sin^2 \theta} \times \frac{\cos^2 \theta \sin\theta}{- \sin^2 \theta}

= -cos3 θ/ sin3 θ 

Given, (x1, y1) = (c/cos θ, c/sin θ)

The equation of tangent is,

y – y1 = m (x – x1)

y - \frac{c}{\sin \theta} = \frac{- \cos^3 \theta}{\sin^3 \theta} \left( x - \frac{c}{\cos \theta} \right)

\frac{y\sin\theta - c}{\sin\theta} = \frac{- \cos^3 \theta}{\sin^3 \theta}\left( \frac{x \cos\theta - c}{\cos\theta} \right)

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)

y - \frac{c}{\sin \theta} = \frac{\sin^3 \theta}{\cos^3 \theta}\left( x - \frac{c}{\cos \theta} \right)

\cos^3 \theta\left( y - \frac{c}{\sin \theta} \right) = \sin^3 \theta\left( x - \frac{c}{\cos \theta} \right)

y \cos^3 \theta - \frac{c \cos^3 \theta}{\sin\theta} = x \sin^3 \theta - \frac{c \sin^3 \theta}{\cos\theta}

x \sin^3 \theta - y \cos^3 \theta = \frac{c \sin^3 \theta}{\cos\theta} - \frac{c \cos^3 \theta}{\sin\theta}

x \sin^3 \theta - y \cos^3 \theta = c\left( \frac{\sin^4 \theta - \cos^4 \theta}{\cos\theta \sin\theta} \right)

x \sin^3 \theta - y \cos^3 \theta = c\left[ \frac{\left( \sin^2 \theta + \cos^2 \theta \right)\left( \sin^2 \theta - \cos^2 \theta \right)}{\cos\theta \sin\theta} \right]

\sin^3 \theta - y \cos^3 \theta =2c \left[ \frac{- \left( \cos^2 \theta - \sin^2 \theta \right)}{2\cos\theta \sin\theta} \right]

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

x\frac{dy}{dx} + y = 0

dy/dx = – y/x

Given, (x1, y1) = (ct, c/t)

Slope of tangent, m= \left( \frac{dy}{dx} \right)_{\left( ct, \frac{c}{t} \right)} =\frac{- \frac{c}{t}}{ct}=\frac{- 1}{t^2}

The equation of tangent is,

y – y1 = m (x – x1)

y - \frac{c}{t} = \frac{- 1}{t^2} \left( x - ct \right)

\frac{yt - c}{t} = \frac{- 1}{t^2} \left( x - ct \right)

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) \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1     at (x1, y1)

Solution:

We have,

\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

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

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

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

dy/dx = – x b2/y a2

Slope of tangent, m= \left( \frac{dy}{dx} \right)_{\left( x_1 , y_1 \right)} =\frac{- x_1 b^2}{y_1 a^2}

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

\frac{{x_1}^2}{a^2} + \frac{{y_1}^2}{b^2} = 1

\frac{{x_1}^2 b^2 + {y_1}^2 a^2}{a^2 b^2} = 1

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

\frac{x x_1}{a^2} + \frac{y y_1}{b^2} = 1

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

\frac{a^2 x}{x_1} - \frac{b^2 y}{y_1} = a^2 - b^2

(xiii) \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1     at (x0 , y0)

Solution:

We have,

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

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

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

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

dy/dx = x b2/y a2

Slope of tangent, m= \left( \frac{dy}{dx} \right)_{\left( x_0 , y_0 \right)} =\frac{x_0 b^2}{y_0 a^2}

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)

\frac{{x_0}^2}{a^2} - \frac{{y_0}^2}{b^2} = 1

x02 b2 – y02 a2 = a2 b

Substituting this in eq(1), we get,

x x0 b2 – y y0 a2 = a2 b2

Dividing this by a2 b2, we get

\frac{x x_0}{a^2} - \frac{y y_0}{b^2} = 1

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

\frac{a^2 x}{x_0} + \frac{b^2 y}{y_0} = a^2 + b^2

(xiv) x^\frac{2}{3} + y^\frac{2}{3} = 2     at (1, 1)

Solution:

We have,

x^\frac{2}{3} + y^\frac{2}{3} = 2

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

\frac{2}{3} x^\frac{- 1}{3} + \frac{2}{3} y^\frac{- 1}{3} \frac{dy}{dx} = 0

\frac{dy}{dx} = \frac{- x^\frac{- 1}{3}}{y^\frac{- 1}{3}} = \frac{- y^\frac{1}{3}}{x^\frac{1}{3}}

Slope of tangent, m= \left( \frac{dy}{dx} \right)_{\left( 1, 1 \right)}     = -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= \left( \frac{dy}{dx} \right)_{\left( 2, 1 \right)}      = 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= \left( \frac{dy}{dx} \right)_{\left( 1, 2 \right)}     = 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 = \left( \frac{dy}{dx} \right)_{\left( 3 \cos\theta, 2 \sin\theta \right)} =\frac{- 12\cos\theta}{18\sin\theta}=\frac{- 2 \cos\theta}{3 \sin\theta}

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 = \left( \frac{dy}{dx} \right)_{\left( x_1 , y_1 \right)} =\frac{2a}{y_1}     = m

The equation of tangent is,

y – y1 = m (x – x1)

y - y_1 = \frac{2a\left( x - x_1 \right)}{y_1}

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) \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1     at (√2a, b)

Solution:

We have,

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

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

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

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

dy/dx = x b2/y a2

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

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

\frac{\sqrt{2}x}{a} - \frac{y}{b} = 1

The equation of normal is, 

y – y1 = -1/m (x – x1)

y – b = – a/√2b(x – √2a)

√2 by – √2 b2 = – ax + √2 a

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

\frac{dx}{d\theta} = 1 + \cos \theta     and \frac{dy}{d\theta} = - \sin \theta

\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{- \sin \theta}{1 + \cos \theta}

Slope of tangent, m = \left( \frac{dy}{dx} \right)_{\theta = \frac{\pi}{4}}

\frac{- \sin \frac{\pi}{4}}{1 + \cos \frac{\pi}{4}}

\frac{\frac{- 1}{\sqrt{2}}}{1 + \frac{1}{\sqrt{2}}}

\frac{-1}{\sqrt{2} + 1}

\frac{-1}{\sqrt{2} + 1}\times\frac{\sqrt{2} - 1}{\sqrt{2} - 1}

= 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

\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}

\frac{- \sin\theta}{1 + \cos\theta}

Slope of tangent, m= \left( \frac{dy}{dx} \right)_{\theta = \frac{\pi}{2}} =\frac{- \sin\frac{\pi}{2}}{1 + \cos\frac{\pi}{2}}

= -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) x = \frac{2 a t^2}{1 + t^2}, y = \frac{2 a t^3}{1 + t^2}     at t = 1/2

Solution:

We have,

x = \frac{2 a t^2}{1 + t^2}, y = \frac{2 a t^3}{1 + t^2}

dx/dt = \frac{\left( 1 + t^2 \right)\left( 4at \right) - 2a t^2 \left( 2t \right)}{\left( 1 + t^2 \right)^2}

\frac{4at}{\left( 1 + t^2 \right)^2}

dy/dt = \frac{\left( 1 + t^2 \right)6a t^2 - 2a t^3 \left( 2t \right)}{\left( 1 + t^2 \right)^2}

 = \frac{6a t^2 + 2a t^4}{\left( 1 + t^2 \right)^2}

\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{6a t^2 + 2a t^4}{\left( 1 + t^2 \right)^2}}{\frac{4at}{\left( 1 + t^2 \right)^2}} = \frac{6a t^2 + 2a t^4}{4at}

Slope of tangent, m= \left( \frac{dy}{dx} \right)_{t = \frac{1}{2}} =\frac{\frac{3a}{2} + \frac{a}{8}}{2a}=\frac{13a}{8}(\frac{1}{2a})=\frac{13}{16}

Given, (x1, y1) = \left(\frac{2 a t^2}{1 + t^2},\frac{2 a t^3}{1 + t^2}\right)=\left( \frac{\frac{a}{2}}{1 + \frac{1}{4}}, \frac{\frac{a}{4}}{1 + \frac{1}{4}} \right) = \left( \frac{\frac{a}{2}}{\frac{5}{4}}, \frac{\frac{a}{4}}{\frac{5}{4}} \right) = \left( \frac{2a}{5}, \frac{a}{5} \right)

The equation of tangent is,

y – y1 = m (x – x1)

y - \frac{a}{5} = \frac{13}{16}\left( x - \frac{2a}{5} \right)

\frac{5y - a}{5} = \frac{13}{16}\left( \frac{5x - 2a}{5} \right)

5y - a = \frac{13}{16}\left( 5x - 2a \right)

80y – 16a = 65x – 26a

65x – 80y – 10a = 0

13x – 16y – 2a = 0

The equation of normal is,

y – y1 = -1/m (x – x1)

y - \frac{a}{5} = \frac{- 16}{13} \left( x - \frac{2a}{5} \right)

\frac{5y - a}{5} = \frac{- 16}{13}\left( \frac{5x - 2a}{5} \right)

5y - a = \frac{- 16}{13}\left( 5x - 2a \right)

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

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

Slope of tangent, m= \left( \frac{dy}{dx} \right)_{t = 1}     = 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

\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{b \sec^2 t}{a \sec t \tan t} = \frac{b}{a}\cosec t

Slope of tangent, m = \left( \frac{dy}{dx} \right)_{t = t} =\frac{b}{a}\cosec t

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)

y - \frac{b \sin t}{\cos t} = \frac{b}{a \sin t}\left( x - \frac{a}{\cos t} \right)

\frac{y \cos t - b \sin t}{\cos t} = \frac{b}{a \sin t}\left( \frac{x \cos t - a}{\cos t} \right)

y \cos t - b \sin t = \frac{b}{a \sin t}\left( x \cos t - a \right)

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)

y - b \frac{\sin t}{\cos t} = \frac{- a}{b}\sin t\left( x - \frac{a}{\cos t} \right)

\frac{y \cos t - b \sin t}{\cos t} = \frac{- a}{b}\sin t\left( \frac{x \cos t - a}{\cos t} \right)

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

\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}

\frac{a\sin\theta}{a\left( 1 + \cos\theta \right)}

\frac{\sin\theta}{\left( 1 + \cos\theta \right)}

\frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2 \cos^2 \frac{\theta}{2}}     

= tan θ/2   . . . . (1)

Slope of tangent, m= \left( \frac{dy}{dx} \right)_\theta =\tan\frac{\theta}{2}

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 - a\left( 2 \sin^2 \frac{\theta}{2} \right) = x\tan\frac{\theta}{2} - a\theta\tan\frac{\theta}{2} - a\frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2 \cos^2 \frac{\theta}{2}}2\sin\frac{\theta}{2}\cos\frac{\theta}{2}

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

\frac{dy}{dx}=\frac{\frac{dy}{d \theta}}{\frac{dx}{d \theta}}

\frac{3\cos \theta-3\sin^2 \theta\cos \theta}{-3\sin \theta+3\cos^2\theta \sin \theta}

\frac{\cos \theta(1-\sin^2 \theta}{-\sin \theta(1-\cos^2 \theta)}

=  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

\frac{dy}{dx} = \frac{4 - 2x}{4y - 6} = \frac{2 - x}{2y - 3}

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


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