RD Sharma Class 12 Ex 16.2 Solutions Chapter 16 Tangents and Normals

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter16
Exercise16.2
CategoryRD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 16.2 Solutions Chapter 16 Tangents and

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 = 6x– 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 = 2x− 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)

sinθ (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 ab2, 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 – xy1 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 – y0a2  . . . . (1)

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

x02 b2 – y02 a2 = ab

Substituting this in eq(1), we get,

x x0 b2 – y y0 a2 = ab2

Dividing this by ab2, 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 – xy0 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 + sinθ)

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 = a+ 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 − cosθ, y = 3 sin θ − sinθ

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 + sinθ = -tanθ (x – 3 cos θ + cos3 θ)

4 (y cosθ – x sinθ) = 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)

sinθ – x cos3 θ = 3sinθ – sin6 θ – 3cosθ + cosθ

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

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 = x+ 2x + 6

dy/dx = 3 x+ 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 – y= 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

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