RD Sharma Class 12 Exercise 15.2 Tangents and Normals Solutions Maths

RD Sharma Class 12 Exercise 15.2 Tangents and Normals Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 21, 2022 01:15 PM IST

RD Sharma Solutions are the highly recommended reference material by most of the CBSE schools. In the absence of a teacher, this book will guide the students in the right direction by clarifying their doubts and providing accurate answers. Mathematics is a subject where doubt arises quite often. And when it comes to chapter 15, things get even more complicated. RD Sharma Solutions The students with such issues can use the RD Sharma Class 12th Exercise 15.2 to clarify their doubts.

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RD Sharma Class 12 Solutions Chapter 15 Tangents and Normals - Other Exercise
Tangents and Normals Excercise: 15.2
RD Sharma Chapter wise Solutions

RD Sharma Class 12 Solutions Chapter 15 Tangents and Normals - Other Exercise

Chapter 15 - Tangents and Normals Ex 15.1

Chapter 15 - Tangents and Normals Ex 15.3

Chapter 15 -Tangents and Normals Ex-FBQ

Chapter 15 -Tangents and Normals Ex-MCQ

Chapter 15 -Tangents and Normals Ex-VSA

Tangents and Normals Excercise: 15.2

Tangents and Normals Exercise 15.2 Question 1

Answer:

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

HINTS:
We find the slope and differentiate slope

\Rightarrow \frac{d y}{d x} = - 1

Given:
$\sqrt{x} + \sqrt{y} = a at the point (\frac{a^{2}}{4}, \frac{a^{2}}{4})$
Solution:
$\begin{aligned} \frac{1}{2 \sqrt{x}} + \frac{1}{2 \sqrt{y}} \cdot \frac{d y}{d x} = 0 \\ \Rightarrow \frac{1}{2 \sqrt{\frac{a^{2}}{4}}} + \frac{1}{2 \sqrt{\frac{a^{2}}{4}}} \cdot \frac{d y}{d x} = 0 \\ \Rightarrow \frac{1}{a} + \frac{1}{a} \cdot \frac{d y}{d x} = 0 \\ \Rightarrow \frac{1}{a} \cdot \frac{d y}{d x} = - \frac{1}{a} \\ \frac{d y}{d x} = - 1 \end{aligned}$
$\begin{aligned} equation of tangent will be (y - y_{1}) = m (x - x_{1}) \\ y - \frac{a^{2}}{4} = - (x - \frac{a^{2}}{4}) \\ x + y = \frac{a^{2}}{2} \end{aligned}$

Tangents and Normals Exercise 15.2 Question 2

Answer: $x + 4 y = 17$
Hint:
We find the slope of tangent and using relation,

slope of normal = \frac{- 1}{slope} of tangent.

Given:

y = 2 x^{3} - x^{2} + 3 at (1, 4)

Solution:
${\frac{d y}{d x} |}_{1, 4]} = {(6 x^{2} - 2 x) |}_{(1, 4)}$
$m = 6 - 2$
$∴ m = 4, slope of tangent$
Equation of normal will be

\Rightarrow y - 4 = - \frac{1}{4} (x - 1)

x + 4 y = 17

Tangents and Normals Exercise 15.2 Question 3 Sub Question 1

ANSWER: Equation of tangent,

y + 10 x - 5 = 0

Equation of normal ,

x - 10 y + 50 = 0

HINTS:
Differentiating the given equation and find the slope of the tangent.
GIVEN:
$y = x^{4} - b x^{3} + 13 x^{2} - 10 x + 5 at (0, 5)$
Solution:
$\frac{d y}{d x} = 4 x^{3} - 3 b x^{2} + 26 x - 10$
$Slope of the tangent$ $m = {(\frac{d y}{d x})}_{(0, 5)} = - 10$
Equation of tangent is,

\begin{aligned} y - y_{1} = m (x - x_{1}) \\ \Rightarrow y - 5 = - 10 (x - 0) \\ \Rightarrow y + 10 x - 5 = 0 \end{aligned}

Equation of Normal is,

\begin{aligned} y - y_{1} = - \frac{1}{m} (x - x_{1}) \\ \Rightarrow y - 5 = \frac{1}{10} (x - 0) \\ \Rightarrow 10 y - 50 = x \\ \Rightarrow x - 10 y + 50 = 0 \end{aligned}

Tangents and Normals Exercise 15.2 Question 3 Sub Question 3

Answer:Equation of tangent,

x = 0

Equation of normal,

y = 0

HINTS:
Differentiating the given equation .
GIVEN:
$y = x^{2} at (0, 0)$
Solution:
$\frac{d y}{d x} = 2 x$
Given $(x, y) = (0, 0)$
Slope of tangent ,

m = {(\frac{d y}{d x})}_{(0, 0)} = 2 (0) = 0

Equation of tangent is,

$\begin{aligned} y - y_{1} = m (x - x_{1}) \\ \Rightarrow y - 0 = 0 (x - 0) \\ \Rightarrow y = 0 \end{aligned}$

Equation of Normal is,

$\begin{aligned} \Rightarrow y - y_{1} = - \frac{1}{m} (x - x_{1}) \\ \Rightarrow y - 0 = - \frac{1}{0} (x - 0) \\ \Rightarrow x = 0 \end{aligned}$

Tangents and Normals Exercise 15.2 Question 3 Sub Question 4

Answer: Equation of tangent,

x - y - 3 = 0

Equation of normal,

x + y + 1 = 0

HINTS:
Differentiating the given equation with respect to x.
Given:
$y = 2 x^{2} - 3 x - 1 at (1, - 2)$
Solution:

\frac{d y}{d x} = 4 x - 3

Slope of tangent,

$m = {(\frac{d y}{d x})}_{| 1, - 2 |} = 4 - 3 = 1$

Equation of tangent is,

$\begin{aligned} y - y_{1} = m (x - x_{1}) \\ \Rightarrow y + 2 = 1 (x - 1) \\ \Rightarrow y + 2 = x - 1 ∴ x - y - 3 = 0 \end{aligned}$

Equation of Normal is,

$\begin{aligned} \Rightarrow y - y_{1} = - \frac{1}{m} (x - x_{1}) \\ \Rightarrow y + 2 = - 1 (x - 1) \\ \Rightarrow y + 2 = - x + 1 \\ \Rightarrow x + y + 1 = 0 \end{aligned}$

Tangents and Normals Exercise 15.2 Question 3 Sub Question 5

Answer: Equation of tangent,

y + 2 x = 2

Equation of normal ,

2 y - x + 6 = 0

HINTS:
Differentiating the given equation with respect to x.
Given:
$y^{2} = \frac{x^{3}}{4 - x} at (2, - 2)$
Solution:
$2 y \frac{d y}{d x} = \frac{(4 - x) 3 x^{2} + x^{3}}{(4 - x)^{2}}$
$\frac{d y}{d x} = \frac{(4 - x) 3 x^{2} + x^{3}}{2 y (4 - x)^{2}}$
$m (tangent) at (2, - 2) = - 2$
$m (normal) at (3, 2) = - \frac{1}{2}$
Equation of tangent is,

\begin{aligned} y_{1} = m (tangent) (x - x_{1}) \\ y + 2 = - 2 (x - 2) \end{aligned}

∴ y + 2 x - 2 = 0

Equation of Normal is,

y_{1} = m (normal) (x - x_{1})

y + 2 = (- \frac{1}{2}) (x - 2)

2 y + 4 = - x + 2

∴ 2 y + x + 2 = 0

Tangents and Normals Exercise 15.2 Question 3 Sub Question 6

ANSWER: Equation of tangent,

10 x - y - 8 = 0

Equation of normal,

x + 10 y - 223 = 0

HINTS:
Differentiating the given curve and find the slope.
GIVEN:
$y = x^{2} + 4 x + 1 at x = 3$
Solution:
$\frac{d y}{d x} = 2 x + 4$
When $x = 3, y = 9 + 12 + 1 = 22$
So, $(x_{1}, y_{1}) = (3, 22)$
Slope of tangent ,

m = {(\frac{d y}{d x})}_{(x = 3)} = 10

Equation of tangent is,

$\begin{aligned} y - y_{1} = m (x - x_{1}) \\ \Rightarrow y - 22 = 10 (x - 3) \\ \Rightarrow y - 22 = 10 x - 30 ∴ 10 x - y - 8 = 0 \end{aligned}$

Equation of Normal is,

$\begin{aligned} \Rightarrow y - y_{1} = - \frac{1}{m} (x - x_{1}) \\ \Rightarrow y - 22 = - \frac{1}{10} (x - 3) \\ \Rightarrow 10 y - 220 = - x + 3 \\ ∴ x + 10 y - 223 = 0 \end{aligned}$

Tangents and Normals Exercise 15.2 Question 3 Sub Question 7

ANSWER: Equation of tangent,

\frac{x}{a} \cos θ + \frac{y}{b} \sin θ = 1

Equation of normal ,

a x \sec θ - b y cosec θ = (a^{2} - b^{2})

HINTS:
Differentiating the given equation with respect to x.
GIVEN:
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 at (a \cos θ, b \sin θ)$
Solution:
$\begin{aligned} \Rightarrow \frac{2 x}{a^{2}} + \frac{2 y}{b^{2}} \cdot \frac{d y}{d x} = 0 \\ \Rightarrow \frac{2 y}{b^{2}} \cdot \frac{d y}{d x} = \frac{- 2 x}{a^{2}} \\ \Rightarrow \frac{d y}{d x} = \frac{- x b^{2}}{y a^{2}} \end{aligned}$
Slope of tangent ,

\begin{aligned} m & = {(\frac{d y}{d x})}_{∣ a \cos θ, b \sin θ]} \\ = \frac{- a \cos θ (b^{2})}{b \sin θ (a^{2})} \\ = \frac{- b \cos θ}{a \sin θ} \end{aligned}

Equation of tangent is,

$\begin{aligned} y - y_{1} = m (x - x_{1}) \\ \Rightarrow y - b \sin θ = \frac{- b \cos θ}{a \sin θ} (x - a \cos θ) \\ \Rightarrow a y \sin θ - a b \sin^{2} θ = - b x \cos θ + a b \cos^{2} θ \\ \Rightarrow b x \cos θ + a y \sin θ = a b \end{aligned}$

Dividing by ab

$∴ \frac{x}{a} \cos θ + \frac{y}{b} \sin θ = 1$

Equation of Normal is,

$\begin{aligned} y - y_{1} = - \frac{1}{m} (x - x_{1}) \\ y - b \sin θ = \frac{a \sin θ}{b \cos θ} (x - a \cos θ) \\ \Rightarrow b y \cos θ - b^{2} \sin θ \cos θ = a x \sin θ - a^{2} \sin θ \cos θ \\ \Rightarrow a x \sin θ - b y \cos θ = (a^{2} - b^{2}) \sin θ \cos θ \end{aligned}$

Dividing by $\sin θ \cos θ$

$∴ a x \sec θ - b y cosec θ = (a^{2} - b^{2})$

Tangents and Normals Exercise 15.2 Question 3 Sub Question 8

ANSWER: Equation of tangent,

\frac{x}{a} \sec θ - \frac{y}{b} \tan θ = 1

Equation of normal ,

a x \cos θ + b y \cot θ = (a^{2} + b^{2})

HINTS:
Differentiating the given curve with respect to x and find its slope.
GIVEN:
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 at (asec θ, b \tan θ)$
SOLUTION:
$\begin{aligned} \Rightarrow \frac{2 x}{a^{2}} - \frac{2 y}{b^{2}} \cdot \frac{d y}{d x} = 0 \\ \Rightarrow \frac{2 y}{b^{2}} \cdot \frac{d y}{d x} = \frac{2 x}{a^{2}} \\ \Rightarrow \frac{d y}{d x} = \frac{x b^{2}}{y a^{2}} \end{aligned}$
Slope of tangent ,

m = {(\frac{d y}{d x})}_{(a \sec θ, b \tan θ)}

= \frac{a \sec θ \cdot b^{2}}{b \tan θ \cdot a^{2}}

= \frac{b}{a} \cdot \frac{1}{\cos θ} \cdot \frac{1}{\frac{\sin θ}{\cos θ}}

[∵ \sec θ = \frac{1}{\cos θ} \cdot \tan θ = \frac{\sin θ}{\cos θ}]

= \frac{b}{a} \cdot \frac{1}{\sin θ} = \frac{b}{a} \cos e c θ

∴ m = {(\frac{d y}{d x})}_{(asec θ, b \tan θ)} = \frac{b}{a \sin θ}

Equation of tangent is,

$\begin{aligned} y - y_{1} = m (x - x_{1}) \\ \Rightarrow y - b \tan θ = \frac{b}{a \sin θ} (x - a \sec θ) \\ \Rightarrow a y \sin θ - a b \frac{\sin θ}{\cos θ} \sin θ = b x - \frac{a b}{\cos θ} \\ \Rightarrow \frac{a y \sin θ \cos θ - a b \sin^{2} θ}{\cos θ} = \frac{b x \cos θ - a b}{\cos θ} \end{aligned}$

$\begin{aligned} \Rightarrow a y \sin θ \cos θ - a b \sin^{2} θ = b x \cos θ - a b \\ \Rightarrow a y \sin θ \cos θ - b x \cos θ = a b \sin^{2} θ - a b \\ \Rightarrow a y \sin θ \cos θ - b x \cos θ = - a b (1 - \sin^{2} θ) \\ \Rightarrow a y \sin θ \cos θ - b x \cos θ = - a b \cos^{2} θ [∵ 1 - \sin^{2} θ = \cos^{2} θ] \end{aligned}$

Divinding by $- a b \cos^{2} θ$

$\begin{aligned} \Rightarrow \frac{a y \sin θ \cos θ - b x \cos θ}{- a b \cos^{2} θ} = 1 \\ \Rightarrow \frac{a y \sin θ \cos θ}{- a b \cos^{2} θ} - \frac{b x \cos θ}{- a b \cos^{2} θ} = 1 \\ \Rightarrow \frac{- y}{b} \frac{\sin θ}{\cos θ} - \frac{x}{a} \frac{1}{\cos θ} = 1 \\ ∴ \frac{x}{a} \sec θ - \frac{y}{b} \tan θ = 1 \end{aligned}$

Equation of Normal is,

$\begin{aligned} y - y_{1} = \frac{1}{m} (x - x_{1}) \\ \Rightarrow y - b \tan θ = \frac{- a \sin θ}{b} (x - asec θ) \\ \Rightarrow a x \sin θ + b y = (a^{2} + b^{2}) \tan θ \end{aligned}$

Dividing by $\tan Θ$

$∴ a x \cos θ + b y \cot θ = (a^{2} + b^{2})$

Tangents and Normals Exercise 15.2 Question 3 Sub Question 9

ANSWER: Equation of tangent,

m^{2} x - m y + a = 0

Equation of normal ,

m^{2} x + m^{3} y - 2 a m^{2} - a = 0

HINTS:
Differentiating the given curve and find its slope.
GIVEN:
$y^{2} = 4 a x at (\frac{a}{m^{2}}, \frac{2 a}{m})$
SOLUTION:
$\begin{aligned} y \frac{d y}{d x} = 2 a \\ \frac{d y}{d x} = \frac{2 a}{y} \end{aligned}$
Equation of tangent is,

\begin{aligned} y - y_{1} = m (x - x_{1}) \\ \Rightarrow y - \frac{2 a}{m} = m (x - \frac{a}{m^{2}}) \\ \Rightarrow \frac{m y - 2 a}{m} = m \frac{m^{2} x - a}{m^{2}} \\ \Rightarrow m y - 2 a = m^{2} x - a \\ \Rightarrow m^{2} x - m y + a = 0 \end{aligned}

Equation of Normal is,

\begin{aligned} \Rightarrow y - y_{1} = \frac{- 1}{slope of tangent} (x - x_{1}) \\ \Rightarrow y - \frac{2 a}{m} = - \frac{1}{m} (x - \frac{a}{m^{2}}) \\ \Rightarrow \frac{m y - 2 a}{m} = - \frac{1}{m} \frac{m^{2} x - a}{m^{2}} \\ \Rightarrow m^{3} y - 2 a m^{2} = - m^{2} x + a \\ \Rightarrow m^{2} x + m^{3} y - 2 a m^{2} - a = 0 \end{aligned}

Tangents and Normals Exercise 15.2 Question 3 Sub Question 10

ANSWER: Equation of tangent,

x \cos^{3} θ + y \sin^{3} θ = c

Equation of normal ,

x \sin^{3} θ - y \cos^{2} θ + 2 c \cot (2 θ) = 0

HINTS:
Differentiating the given curve with respect to x and find its slope.

\sin^{2} θ + \cos^{2} θ = 1

GIVEN:
$c^{2} (x^{2} + y^{2}) = x^{2} y^{2} a t [\frac{c}{\cos θ}, \frac{c}{\sin θ}]$
SOLUTION:
$\begin{aligned} \Rightarrow 2 x c^{2} + 2 y c^{2} \frac{d y}{d x} = x^{2} 2 y \frac{d y}{d x} + 2 x y^{2} \\ \Rightarrow \frac{d y}{d x} (2 y c^{2} - 2 x^{2} y) = 2 x y^{2} - 2 x c^{2} \\ \Rightarrow \frac{d y}{d x} = \frac{x y^{2} - x c^{2}}{y c^{2} - x^{2} y} \end{aligned}$
Slope of tangent,

m = {(\frac{d y}{d x})}_{(\frac{c}{\cos θ}, \frac{c}{\sin θ})}

= \frac{\frac{c^{3}}{\cos θ \sin^{2} θ} - \frac{c^{3}}{\cos θ}}{\frac{c^{3}}{\sin θ} - \frac{c^{3}}{\cos^{2} θ \sin θ}}

= \frac{\frac{1 - \sin^{2} θ}{\cos θ \sin^{2} θ}}{\frac{\cos^{2} θ - 1}{\cos^{2} θ \sin θ}}

= \frac{\cos^{2} θ}{\cos θ \sin^{2} θ} \times \frac{\cos^{2} θ \sin θ}{- \sin^{2} θ}

= \frac{\cos^{3} θ}{- \sin^{3} θ}

Given,

(x_{1}, y_{1}) = (\frac{c}{\cos θ}, \frac{c}{\sin θ})

Equation of tangent is,

$\begin{aligned} y - y_{1} = m (x - x_{1}) \\ \Rightarrow y - \frac{c}{\sin θ} = \frac{- \cos^{3} θ}{\sin^{3} θ} (x - \frac{c}{\cos θ}) \\ \Rightarrow \frac{y \sin θ - c}{\sin θ} = \frac{- \cos^{3} θ}{\sin^{3} θ} (\frac{x \cos θ - c}{\cos θ}) \\ \Rightarrow \sin^{2} θ (y \sin θ - c) = - (\cos^{2} θ x \cos θ - c) \\ \Rightarrow y \sin^{3} θ - c \sin^{2} θ = - x \cos^{3} θ + c \cos^{2} θ \\ \Rightarrow x \cos^{3} θ + y \sin^{3} θ = c (\sin^{2} θ + \cos^{2} θ) \\ \Rightarrow x \cos^{3} θ + y \sin^{3} θ = c \end{aligned}$
Equation of Normal is,
$\begin{aligned} y - y_{1} = - \frac{1}{m} (x - x_{1}) \\ \Rightarrow y - \frac{c}{\sin θ} = \frac{\sin^{3} θ}{\cos^{3} θ} (x - \frac{c}{\cos θ}) \\ \Rightarrow \cos^{3} θ (y - \frac{c}{\sin θ}) = \sin^{3} θ (x - \frac{c}{\cos θ}) \\ \Rightarrow y \cos^{3} θ - c \frac{\cos^{3} θ}{\sin θ} = x \sin^{3} θ - c \frac{\sin^{3} θ}{\cos θ} \end{aligned}$
$\begin{aligned} \Rightarrow x \sin^{3} θ - y \cos^{3} θ = c (\frac{\sin^{4} θ - \cos^{4} θ}{\cos θ \sin θ}) \\ \Rightarrow x \sin^{3} θ - y \cos^{3} θ = c [\frac{(\sin^{2} θ + \cos^{2} θ) (\sin^{2} θ - \cos^{2} θ)}{\cos θ \sin θ}] \\ \Rightarrow x \sin^{3} θ - y \cos^{3} θ = 2 c [\frac{- (\cos^{2} θ - \sin^{2} θ)}{2 \cos θ \sin θ}] \end{aligned}$
$\begin{aligned} \Rightarrow x \sin^{3} θ - y \cos^{3} θ = 2 c [\frac{- \cos (2 θ)}{\sin (2 θ)}] \\ \Rightarrow x \sin^{3} θ - y \cos^{3} θ = - 2 c \cot (2 θ) \\ \Rightarrow x \sin^{3} θ - y \cos^{3} θ + 2 c \cot (2 θ) = 0 \end{aligned}$

Tangents and Normals Exercise 15.2 Question 3 Sub Question 11

ANSWER: Equation of tangent,

x + y t^{2} = 2 c t

Equation of normal ,

x t^{3} - y t = c t^{4} - c

HINTS:
Differentiating the given curve with respect to x and find its slope.
GIVEN:
$x y = c^{2} at (c t, \frac{c}{t})$
SOLUTION:
$\begin{aligned} x \frac{d y}{d x} + y = 0 \\ \Rightarrow \frac{d y}{d x} = - \frac{y}{x} \end{aligned}$
Slope of tangent,

m = (\frac{d y}{d x}) (ct, \frac{c}{t}) = \frac{- \frac{c}{t}}{c t} = - \frac{1}{t^{2}}

Equation of tangent is,

$\begin{aligned} y - y_{1} = m (x - x_{1}) \\ \Rightarrow y - \frac{c}{t} = - \frac{1}{t^{2}} (x - c t) \\ \Rightarrow \frac{y t - c}{t} = - \frac{1}{t^{2}} (x - c t) \\ \Rightarrow y t^{2} - c t = - x + c t \\ ∴ x + y t^{2} = 2 c t \end{aligned}$
Equation of Normal is,

$\begin{aligned} \Rightarrow y - y_{1} = - \frac{1}{m} (x - x_{1}) \\ \Rightarrow y - \frac{c}{t} = t^{2} (x - c t) \\ \Rightarrow y t - c = t^{3} x - c t^{4} \\ \Rightarrow x t^{3} - y t = c t^{4} - c \end{aligned}$

Tangents and Normals Exercise 15.2 Question 3 Sub Question 12

ANSWER: Equation of tangent,

\frac{x x_{1}}{a_{2}} + \frac{y y_{1}}{a_{2}} = 1

Equation of normal,

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

HINTS:
Differentiating the given equation
GIVEN:

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

.....(1)
SOLUTION:
Since

P (x_{1}, y_{1})

lies on the curve(i)

\frac{x_{1}^{2}}{a^{2}} + \frac{y_{1}^{2}}{b^{2}} = 1

.......(ii)

\frac{2 x}{a^{2}} + \frac{2 y}{b^{2}} \cdot \frac{d y}{d x} = 0

\Rightarrow \frac{d y}{d x} = - \frac{b^{2} x_{1}}{a^{2} y_{1}}

Equation of tangent at

P (x_{1}, y_{1})

\begin{aligned} (y - y_{1}) = {(\frac{d y}{d x})}_{(x_{1}, y_{1})} (x - x_{1}) \\ \Rightarrow (y - y_{1}) = - \frac{b^{2} x_{1}}{a^{2} y_{1}} (x - x_{1}) \\ \Rightarrow \frac{y y_{1} - y_{1}^{2}}{b^{2}} = (\frac{- x x_{1} + x_{1}^{2}}{a^{2}}) \end{aligned}

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

[U s i n g (i)]

Equation of Normal at

P (x_{1}, y_{1})

is,

\Rightarrow y - y_{1} = - \frac{1}{{[\frac{d y}{d x}]}_{∣ x_{1}, y_{1}}}} (x - x_{1})

\Rightarrow \frac{b^{2} (y - y_{1})}{y_{1}} = \frac{a^{2} (x - x_{1})}{x_{1}}

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

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

[U s i n g (i i)]

Tangents and Normals Exercise 15.2 Question 3 Sub Question 13

ANSWER: Equation of tangent,

\frac{x x_{0}}{a^{2}} - \frac{y y_{0}}{b^{2}} = 1

Equation of normal ,

y - y_{0} = \frac{α^{2}}{b^{2}} \cdot \frac{y_{0}}{x_{0}} (x - x_{0})

HINTS:
Differentiating the given curve with respect to x and find its slope.
GIVEN:
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 at (x_{0}, y_{0})$ ..............(i)
SOLUTION:
Since

P (x_{0}, y_{0})

lies on the curve(i)

\begin{aligned} \frac{2 x}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \\ \frac{2 x}{a^{2}} - \frac{2 y}{b^{2}} \cdot \frac{d y}{d x} = 0 \\ \Rightarrow \frac{d y}{d x} = \frac{x b^{2}}{y a^{2}} \end{aligned}

Slope of tangent,

${\Rightarrow \frac{d y}{d x} |}_{i x_{0}, y_{0}]} = \frac{x_{0} b^{2}}{y_{0} a^{2}} = m$
Equation of tangent is

$\begin{aligned} y - y_{0} = m (x - x_{0}) \\ \Rightarrow y - y_{0} = \frac{x_{0} b^{2}}{y_{0} a^{2}} (x - x_{0}) \end{aligned}$

$\Rightarrow \frac{y_{0}}{b^{2}} (y - y_{0}) = \frac{x^{0}}{a^{2}} (x - x_{0})$

$\begin{aligned} \Rightarrow \frac{y y_{0}}{b^{2}} - \frac{y_{0}^{2}}{b^{2}} = \frac{x_{0} x}{a^{2}} - \frac{x_{0}^{2}}{a^{2}} \\ \Rightarrow \frac{y y_{0}}{b^{2}} - \frac{x x_{0}}{a^{2}} = \frac{y_{0}^{2}}{b^{2}} - \frac{x_{0}^{2}}{a^{2}} \end{aligned}$

$\Rightarrow \frac{x x_{0}}{a^{2}} - \frac{y y_{o}}{b^{2}} = 1$
Again, slope of normal,

$\begin{aligned} m_{1} m_{2} = - 1 \\ \frac{b^{2}}{a^{2}} \cdot \frac{x_{0}}{y_{0}} \cdot m_{2} = - 1 \\ m_{2} = - \frac{a^{2}}{b^{2}} \frac{y^{0}}{x^{0}} a t (x_{0}, y_{0}) \end{aligned}$
Equation of Normal is ,

$∴ y - y_{0} = - \frac{a^{2}}{b^{2}} \frac{y^{0}}{x^{0}} (x - x_{0})$

$\Rightarrow \frac{b^{2} (y - y_{0})}{y_{0}} = - \frac{a^{2} (x - x_{0})}{x_{0}}$

$\Rightarrow \frac{b^{2} y}{y_{0}} - b^{2} = - \frac{a^{2} x}{x_{0}} + a^{2}$

$\Rightarrow \frac{a^{2} x}{x_{0}} + \frac{b^{2} y}{y_{0}} = a^{2} + b^{2}$

Tangents and Normals Exercise 15.2 Question 3 Sub Question 14

ANSWER: Equation of tangent,

x + y - 2 = 0

Equation of normal ,

y - x = 0

HINTS:
Differentiating the given curve with respect to x and find its slope first.
GIVEN:
$x^{\frac{2}{3}} + y^{\frac{2}{3}} = 2 at (1, 1)$
SOLUTION:
$\begin{aligned} \frac{2}{3} x^{\frac{- 1}{3}} + \frac{2}{3} y^{\frac{- 1}{3}} \frac{d y}{d x} = 0 \\ \Rightarrow \frac{d y}{d x} = - \frac{x^{\frac{- 1}{3}}}{y^{\frac{- 1}{3}}} = \frac{- y^{\frac{1}{3}}}{x^{\frac{1}{3}}} \end{aligned}$
Slope of tangent,

m = {(\frac{d y}{d x})}_{(1, 1)} = \frac{- 1}{1} = - 1

Equation of tangent is

$\begin{aligned} y - y_{1} = m (x - x_{1}) \\ \Rightarrow y - 1 = - 1 (x - 1) \\ \Rightarrow y - 1 = - x + 1 \\ ∴ x + y - 2 = 0 \end{aligned}$
Equation of Normal is ,

$\begin{aligned} y - y_{1} = - \frac{1}{m} (x - x_{1}) \\ \Rightarrow y - 1 = 1 (x - 1) \\ \Rightarrow y - 1 = x - 1 \\ ∴ y - x = 0 \end{aligned}$

Tangents and Normals Exercise 15.2 Question 3 Sub Question 15

ANSWER: Equation of tangent,

x - y - 1 = 0

Equation of normal ,

x + y - 3 = 0

HINTS:
Differentiating the given curve with respect to x and find its slope .
GIVEN:

x^{2} = 4 y at (2, 1)

SOLUTION:
$\begin{aligned} 2 x = 4 \frac{d y}{d x} \\ \Rightarrow \frac{d y}{d x} = \frac{x}{2} \end{aligned}$
Slope of tangent,

m = {(\frac{d y}{d x})}_{(2, 1)} = \frac{2}{2} = 1

Equation of tangent is

\begin{aligned} y - y_{1} = m (x - x_{1}) \\ \Rightarrow y - 1 = - 1 (- x + 2) \\ \Rightarrow y - 1 = x - 2 \\ ∴ x - y - 1 = 0 \end{aligned}

Equation of Normal is ,

\begin{aligned} y - y_{1} = - \frac{1}{m} (x - x_{1}) \\ \Rightarrow y - 1 = 1 (- x + 2) \\ \Rightarrow y - 1 = - x + 2 \\ ∴ x + y - 3 = 0 \end{aligned}

Tangents and Normals Exercise 15.2 Question 3 Sub Question 16

ANSWER: Equation of tangent,

x - y + 1 = 0

Equation of normal ,

y + x = 3

HINTS:
Differentiating the given curve with respect to x .
GIVEN:

y^{2} = 4 x at (1, 2)

SOLUTION:

$\begin{aligned} 2 y \frac{d y}{d x} & = 4 \\ \frac{d y}{d x_{(1, 2)}} & = \frac{4}{2 y} \\ = \frac{2}{y} = 1 \end{aligned}$

Slope of tangent = $\frac{d y}{d x} = 1$
Slope of normal = $- \frac{1}{slope of tangent} = - 1$
Equation of tangent is
$\begin{aligned} y - 2 = 1 (x - 1) \\ ∴ x - y + 1 = 0 \end{aligned}$
Equation of Normal is ,

$\begin{aligned} y - 2 = - 1 (x - 1) \\ ∴ y + x = 3 \end{aligned}$

Tangents and Normals Exercise 15.2 Question 3 Sub Question 17

ANSWER: Equation of tangent,

2 x \cos θ + 3 y \sin θ = 6

Equation of normal ,

3 x \sin θ - 2 y \cos θ - 5 \sin θ \cos θ = 0

HINTS:
Differentiating the given curve with respect to x
GIVEN:

4 x^{2} + 9 y^{2} = 36 a t [3 \cos θ, 2 \sin θ]

SOLUTION:
$\begin{aligned} 8 x + 18 y \frac{d y}{d x} = 0 \\ \Rightarrow 18 y \frac{d y}{d x} = - 8 x \\ \Rightarrow \frac{d y}{d x} = \frac{- 8 x}{18 y} = \frac{- 4 x}{9 y} \end{aligned}$
Slope of tangent ,

m = {(\frac{d y}{d x})}_{(3 \cos θ, 2 \sin θ)}

Equation of tangent is,

$y - y_{1} = m (x - x_{1})$

$\Rightarrow y - 2 \sin θ = \frac{- 2 \cos θ}{3 \sin θ} (x - 3 \cos θ)$

$\begin{aligned} \Rightarrow 3 y \sin θ - 6 \sin^{2} θ = - 2 x \cos θ + 6 \cos^{2} θ \\ \Rightarrow 2 x \cos θ + 3 y \sin θ = 6 (\cos^{2} θ + \sin^{2} θ) \end{aligned}$

$∴ 2 x \cos θ + 3 y \sin θ = 6$

Equation of Normal is,

$y - y_{1} = - \frac{1}{m} (x - x_{1})$

$\Rightarrow y - 2 \sin θ = \frac{3 \sin θ}{2 \cos θ} (x - 3 \cos θ)$

$\begin{aligned} \Rightarrow 2 y \cos θ - 4 \sin θ \cos θ = 3 x \sin θ - 9 \sin θ \cos θ \\ \Rightarrow 3 x \sin θ - 2 y \cos θ - 5 \sin θ \cos θ = 0 \end{aligned}$

Tangents and Normals Exercise 15.2 Question 3 Sub Question 18

ANSWER: Equation of tangent,

y y_{1} = 2 a (x + x_{1})

Equation of normal ,

y - y_{1} = \frac{- y_{1}}{2 a} (x - x_{1})

HINTS:
Differentiating with respect to x and find its slope.
GIVEN:

y^{2} = 4 a x at (x_{1}, y_{1})

SOLUTION:

$\begin{aligned} 2 y \frac{d y}{d x} = 4 a \\ \Rightarrow \frac{d y}{d x} = \frac{2 a}{y} \end{aligned}$

$Slope of tangent = \frac{d y}{d x_{∣ x_{1}, y_{1}}}} = \frac{2 a}{y_{1}} = m$
Equation of tangent is

$\begin{aligned} y - y_{1} = m (x - x_{1}) \\ \Rightarrow y - y_{1} = \frac{2 a (x - x_{1})}{y_{1}} \end{aligned}$

$\begin{aligned} \Rightarrow y y_{1} - y_{1}^{2} = 2 a x - 2 a x_{1} \\ \Rightarrow y y_{1} - 4 a x_{1} = 2 a x - 2 a x_{1} \\ \Rightarrow y y_{1} = 2 a x + 2 a x_{1} \\ ∴ y y_{1} = 2 a (x + x_{1}) \end{aligned}$
Equation of Normal is ,
$y - y_{1} = - \frac{1}{slope of tangent} (x - x_{1})$
$y - y_{1} = - \frac{1}{m} (x - x_{1})$
$\Rightarrow y - y_{1} = \frac{- y_{1}}{2 a} (x - x_{1})$

Tangents and Normals Exercise 15.2 Question 4

ANSWER: $(\sqrt{2} - 1) x + y + (\frac{π}{4}) (1 - \sqrt{2}) = 0$
HINTS:
To find equation, first find slope and one point on the tangent.
GIVEN:

x = θ + \sin θ, y = 1 + \cos θ at θ = \frac{π}{4}

SOLUTION:
$\begin{aligned} ∵ Point on the tangent at θ = \frac{π}{4} \\ \Rightarrow x = \frac{π}{4} + \sin (\frac{π}{4}) = \frac{π}{4} + \frac{1}{\sqrt{2}} and y = 1 + \cos \frac{π}{4} = 1 + \frac{1}{\sqrt{2}} \\ point P (\frac{π}{4} + \frac{1}{\sqrt{2}}, 1 + \frac{1}{\sqrt{2}}) lies on the line \end{aligned}$
To find two slope, we find

\frac{d y}{d x}

\frac{d x}{d θ} = 1 + \cos θ and \frac{d y}{d θ} = - \sin θ

Thus

\frac{d y}{d x} = - \frac{\sin θ}{1 + \cos θ}

Solving above

\frac{d y}{d x} = \frac{- 2 \sin \frac{θ}{2} \cos \frac{θ}{2}}{2 \cos^{2} \frac{θ}{2}} or \frac{d y}{d x} = - \tan \frac{θ}{2}

Thus

\frac{d y}{d x} = - \tan (\frac{θ}{2}) at θ = \frac{π}{4}

Slope,

\Rightarrow m = - \tan (\frac{π}{8}) - (1 - \sqrt{2)}

Now apply point slope form to get equation of tangent,

[y - (1 + \frac{1}{\sqrt{2}})] = (1 - \sqrt{2}) (x - \frac{π}{4} - \frac{1}{\sqrt{2}})

\Rightarrow (\sqrt{2} - 1) x + y - 1 - \frac{1}{\sqrt{2}} + (\frac{π}{4} + \frac{1}{\sqrt{2}}) (1 - \sqrt{2}) = 0

Tangents and Normals Exercise 15.2 Question 5 Sub Question 1

ANSWER: Equation of tangent,

2 x + 2 y - π - 4 = 0

Equation of normal ,

2 x - 2 y = π

HINTS:
Differentiate the given equation with respect to and to get the slope of the tangent.
GIVEN:

x = θ + \sin θ, y = 1 + \cos θ at θ = \frac{π}{2}

SOLUTION:
Given as

x = θ + \sin θ, y = 1 + \cos θ at θ = \frac{π}{2}

On differentiating ,

\frac{d x}{d θ} = 1 + \cos θ, \frac{d y}{d θ} = - \sin θ

∴ \frac{d y}{d x} = - \frac{\sin θ}{1 + \cos θ}

m (tangent) at θ = (\frac{π}{2}) - 1

The normal is perpendicular to tangent, therefore ,

m_{1} m_{2} = - 1

m (normal) at θ - (\frac{π}{2}) - 1

The equation of tangent is given by,

\begin{aligned} y - y_{1} = m (tangent) [x - x_{1}] \\ \Rightarrow y - 1 = - 1 [x - (\frac{π}{2}) - 1] \end{aligned}

\begin{aligned} \Rightarrow 2 (y - 1) + 2 x - π - 2 = 0 \\ \Rightarrow & 2 x + 2 y - π - 4 \end{aligned}

The equation of Normal is given by ,

\begin{aligned} y - y_{1} = m (normal) [x - x_{1}] \\ \Rightarrow y - 1 = 1 [x - (\frac{π}{2}) - 1] \end{aligned}

2 x - 2 y = π

Tangents and Normals Exercise 15.2 Question 5 Sub Question 2

ANSWER: Equation of tangent,

y - (\frac{4}{5}) = (\frac{13}{16}) [x - (\frac{2 a}{5})]

Equation of normal ,

y - (\frac{4}{5}) = (\frac{- 16}{13}) [x - (\frac{2 a}{5})]

HINTS:
Differentiate the given equation with respect to t and to get the slopes.
GIVEN:

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

SOLUTION:
Upon differentiation,

\begin{aligned} \frac{d x}{d t} = \frac{(1 + t^{2}) y a t - 2 a t^{2} (2 t)}{{(1 + t^{2})}^{2}} \Rightarrow \frac{d x}{d t} = \frac{4 a t}{{(1 + t^{2})}^{2}} \\ \Rightarrow \frac{d y}{d t} = \frac{(1 + t^{2}) 6 a t^{2} - 2 a t^{2} (2 t)}{{(1 + t^{2})}^{2}} \Rightarrow \frac{d x}{d t} = \frac{6 a t^{2} + 4 a t^{4}}{{(1 + t^{2})}^{2}} \\ ∴ \frac{d y}{d x} = \frac{(6 a t^{2} + 2 a t^{4})}{4 a t} \end{aligned}

m (tangent) at t = \frac{1}{2} is \frac{13}{16}

The normal is perpendicular to tangent, therefore ,

m_{1} m_{2} = - 1

m (normal) at t = (\frac{1}{2}) is - \frac{16}{13}

The equation of tangent is given by,

\begin{aligned} y - y_{1} = m (tangent) [x - x_{1}] \\ \Rightarrow y - [\frac{4}{5}] = (\frac{13}{16}) [x - (\frac{2 a}{5})] \end{aligned}

The equation of Normal is given by ,

\begin{aligned} y - y_{1} = m (normal) [x - x_{1}] \\ \Rightarrow y - [\frac{4}{5}] = (\frac{- 16}{13}) [x - (\frac{2 a}{5})] \end{aligned}

Tangents and Normals Exercise 15.2 Question 6

ANSWER: The equation of normal x=2
HINTS:
Differentiating with respect to x to get its slope.
GIVEN:

x^{2} + 2 y^{2} - 4 x - 6 y + 8 = 0 a t x = 2

SOLUTION:
Upon differentiation

\begin{aligned} 2 x + [4 y [\frac{d y}{d x}]] - 4 - 6 [\frac{d y}{d x}] = 0 \\ \Rightarrow \frac{d y}{d x} = \frac{(4 - 2 x)}{(4 y - 6)} \end{aligned}

Finding the y coordinate by substitute x in the given curve

\Rightarrow 2 y^{2} - 6 y + 4 = 0

\Rightarrow y^{2} - 3 y + 2 = 0

y = 2

y = 1

The normal is perpendicular to tangent, therefore ,

m_{1,} m_{2} = - 1

m (n o r m a l) a t x = 2 i s \frac{1}{0}

Which is Undefined
The equation of Normal is given by ,

\begin{aligned} y - y_{1} = \frac{- 1}{m} (x - x_{1}) \\ \Rightarrow y - 1 = \frac{- 1}{0} (x - 2) \\ \Rightarrow - (x - 2) = 0 \\ \Rightarrow x - 2 = 0 \\ \Rightarrow x = 2 \end{aligned}

when y=2
Slope of tangent =

(\frac{d y}{d x}) (_{2, 1}) = \frac{0}{- 1} = 0

Equation of normal is

\begin{aligned} y - y_{1} = \frac{- 1}{m} (x - x_{1}) \\ \Rightarrow y - 2 = \frac{- 1}{0} (x - 2) \\ \Rightarrow - (x - 2) = 0 \\ \Rightarrow x - 2 = 0 \\ \Rightarrow x = 2 \end{aligned}

In both cases the equation of normal is x=2

Tangents and Normals Exercise 15.2 Question 5 Sub Question 3

Answer:: Equation of tangent,

y - 2 a = 1 (x - a)

Equation of normal ,

y - 2 a = 1 (x - a)

Hint:
Differentiating with respect to x to get its slope.
Given: $x - a t^{2}, y = 4 a t a t t = 1$
Solution:
Upon differentiation

\frac{d x}{d t} = 2 a t, \frac{d y}{d t} = 2 a

∴ \frac{d y}{d t} = \frac{1}{t}

M(tangent) at t=1 is 1
The normal is perpendicular to tangent, therefore ,

m_{1} m_{2} = - 1

m (N o r m a l) a t t = 1 i s - 1

The equation of tangent is given by,

y - y_{1} = m (t a n g e n t) [x - x_{1}]

\Rightarrow y - 2 a = 1 (x - a)

The equation of Normal is given by ,

y - y_{1} = m (t a n g e n t) [x - x_{1}]

\Rightarrow y - 2 a = - 1 (x - a)

Tangents and Normals Exercise 15.2 Question 5 Sub Question 4

ANSWER: Equation of tangent,

y - b \tan t = [\frac{b cosect}{a}] [x - a \sec t]

Equation of normal ,

y - b \tan t = [\frac{- a \sin t}{b}] [x - a \sec t]

HINTS:
Differentiating with respect to x to get its slope.
GIVEN:

x = a s e c t, y = b t a n t a t t

SOLUTION:
Upon differentiation

\begin{aligned} \frac{d x}{d t} = a \sec t \tan t, \frac{d y}{d t} = b \sec^{2} t \\ ∴ \frac{d y}{d x} = \frac{b cosect}{a} \\ M (tangent) at \end{aligned}

The normal is perpendicular to tangent, therefore ,

m_{1}, m_{2} = - 1

m (n o r m a l) a t t = (- \frac{3}{6}) \sin t

The equation of tangent is given by,

y - b \tan t = \frac{b cossect}{a} (x - a \sec t)

The equation of Normal is given by ,

y - b \tan t = \frac{- a \sin t}{b} [x - a \sec t]

Tangents and Normals Exercise 15.2 Question 5 Sub Question 6

ANSWER: Equation of tangent,

y - 3 \sin θ + \sin^{3} θ = - \tan^{3} θ (x - 3 \cos θ + 3 \cos^{3} θ)

Equation of normal ,

y - 3 \sin θ + \sin^{3} θ = \cot^{3} θ (x - 3 \cos θ + 3 \cos^{3} θ)

HINTS:
Differentiating with respect to

θ

to get its slope.
GIVEN:

x = 3 \cos θ - \cos^{2} θ, y = 3 \sin θ - \sin^{3} θ

SOLUTION:
Upon differentiation

\begin{aligned} \frac{d x}{d θ} = 3 \cos θ + 3 \cos^{2} θ \sin θ, \frac{d x}{d θ} = 3 \cos θ - 3 \sin^{2} θ \cos θ \\ ∴ \frac{d y}{d x} = \frac{3 \cos θ - 3 \sin^{2} θ \cos θ}{- 3 \sin θ + 3 \cos^{2} \sin θ} = - \tan^{3} θ \\ m (tangent) at θ is - \tan^{3} θ \end{aligned}

The normal is perpendicular to tangent, therefore ,

m_{1}, m_{2} = - 1

m(normal0 at

θ

\cot^{3} θ

The equation of tangent is given by,

\begin{aligned} y - y_{1} = m (tangent) (x - x_{1}) \\ \Rightarrow y - 3 \sin θ + \sin^{3} θ = - \tan^{3} θ (x - 3 \cos θ + 3 \cos^{3} θ) \end{aligned}

The equation of Normal is given by ,

\begin{aligned} y - y_{1} = m (normal) (x - x_{1}) \\ \Rightarrow y - 3 \sin θ + \sin^{3} θ = \cot^{3} θ (x - 3 \cos θ + 3 \cos^{3} θ) \end{aligned}

Tangents and Normals Exercise 15.2 Question 7

ANSWER: The equation of normal

y - a m^{3} = (\frac{- 2}{3 m}) (x - a m^{2})

HINTS:
Differentiating with respect to x to get its slope of tangent

2 a y (\frac{d y}{d x}) = 3 x^{2}

GIVEN:

a y^{2} = x^{3} a t t h e p o i n t (a m^{2}, a m^{2})

SOLUTION:
Upon differentiation

\frac{d y}{d x} = \frac{3 x^{2}}{2 a y}

m(tangent) at

(a m^{2}, a m^{3}) i s \frac{3 m}{2}

The normal is perpendicular to tangent, therefore ,

m_{1}, m_{2} = - 1

m(tangent) at at

(a m^{2}, a m^{3}) i s \frac{- 2}{3 m}

The equation of Normal is given by ,

\Rightarrow y - a m^{3} = (- \frac{2}{3 m}) (x - a m^{2})

Tangents and Normals Exercise 15.2 Question 8

ANSWER:

a = 2, b = - 7

HINTS:
Differentiating the curve to get its slope of tangent
GIVEN:

y^{2} = a x^{3} + b i s y = 4 x - 5

SOLUTION:
Upon differentiation

\begin{aligned} 2 y (\frac{d y}{d x}) = 3 a x^{2} \\ \Rightarrow \frac{d y}{d x} = \frac{3 a x^{2}}{2 y} \\ m (tangent) at (2, 3) = 2 a \end{aligned}

The equation of tangent is given by

y - y_{1} = m (t a n g e n t) [x - x_{1}]

Comparing the slopes of a tangent with the given equation

2 a = 4

a = 2

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

3^{2} = 2 \times 2^{3} + b

b = - 7

Tangents and Normals Exercise 15.2 Question 9

ANSWER: Equation of tangent:

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

HINTS:
Differentiate with respect to x to get its slope
GIVEN:

y = x^{2} + 4 x - 16

Which is parallel to

3 x - y + 1 = 0

SOLUTION:
Upon differentiation

\frac{d y}{d x} = 2 x + 4 \Rightarrow m (t a n g e n t) = 2 x + 4

The equation of tangent is given by

y - y_{1} = m (t a n g e n t) [x - x_{1}]

Comparing the slopes of a tangent with the given equation

2 x + y = 3 \Rightarrow x = - \frac{1}{2}

Substitutes the value of x in the curve to find y

y = (\frac{1}{4}) - 2 - 16 = - \frac{71}{4}

So,the equation of tangent is parallel to the given line is

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

Tangents and Normals Exercise 15.2 Question 10

ANSWER: Equation of normal:

y - 18 = (- \frac{1}{14}) (x - 2) or y + 6 = (- \frac{1}{14}) (x - 2)

HINTS:
Differentiate the given curve, to get slope of tangent
GIVEN:

y = x^{3} + 2 x + 6

Which is parallel to

x + 14 y + 4 = 0

SOLUTION:
Upon differentiation

\frac{d y}{d x} = 3 x^{2} + 2 \Rightarrow m (t a n g e n t) = 3 x^{2} + 2

The normal is perpendicular to tangent, therefore ,

m_{1}, m_{2} = - 1

m(normal) at

- \frac{1}{3 x^{2} + 2}

The equation of Normal is given by ,

y - y_{1} = m (t a n g e n t) (x - x_{1})

On comparing the slope of normal with the given equation
m(normal)

= - \frac{1}{14}

- \frac{1}{14} = - \frac{1}{(3 x^{2} + 2)} \Rightarrow x = 2 o r - 2

Thus ,the corresponding value of y is 18 or -6
Therefore ,the equation of normal are

y - 18 = (- \frac{1}{14}) (x - 2) or y + 6 = (- \frac{1}{14}) (x + 2)

Tangents and Normals Exercise 15.2 Question 11

ANSWER: Equation of tangent:

9 x - y - 3 = 0 o r 9 x - y + 13 = 0

HINTS:
Slope of tangent = slope of perpendicular line
GIVEN:

y = 4 x^{3} - 3 x + 5

Which is perpendicular to

9 y + x + 3 = 0

SOLUTION:
Let

(x_{1}, y_{1})

be a point on the curve which are used to find the tangents
Slope of the given line =

- \frac{1}{9}

Since tangent is perpendicular to the given line ,
Slope of the tangent=

\frac{- 1}{(- \frac{1}{9})} = 9

Let

(x_{1}, y_{1})

be a point where the tangent is drawn to this curve
Since, the point lies on the curve
Hence,

y_{1} = 4 x_{1}^{3} - 3 x_{1} + 5

Slope of the tangent =slope of the perpendicular line

\begin{aligned} \Rightarrow 12 x_{1}^{2} - 3 = 9 \Rightarrow 12 x_{1}^{2} = 12 \\ \Rightarrow x_{1}^{2} = 1 \Rightarrow x_{1} = \pm 1 \\ Case- 1 : x_{1} = 1 \\ y_{1} = 4 x_{1}^{3} - 3 x_{1} + 5 = 4 - 3 + 5 = 6 \\ ∴ (x_{1}, y_{1}) = (1, 6) \\ Case- 2 : x_{1} = - 1 \\ y_{1} = 4 x_{1}^{3} - 3 x_{1} + 5 = - 4 + 3 + 5 = 4 \\ ∴ (x_{1}, y_{1}) = (- 1, 4) \end{aligned}

Thus, the equation of tangent is (y - 6) = 9 (x - 1) \Rightarrow 9 x - y - 3 = 0 (y - 4) = 9 (x + 1) \Rightarrow 9 x - y + 13 = 0

Tangents and Normals Exercise 15.2 Question 12

ANSWER: $e^{2} (x - y) = 3$
HINTS:
First find the coordinates
GIVEN:
$y = x l o g_{e} x$ Which is parallel to $2 x - 2 y + 3 = 0$
SOLUTION:
$2 x - 2 y + 3 = 0$ $Y = x l o g_{e} x$
$m = 1$ Upon differentiation

\begin{aligned} \frac{d y}{d x} = x \cdot \frac{1}{x} + \log x \cdot 1 \\ \Rightarrow 1 + \log x = - 1 \\ ∴ \log_{e} x = - 2 \\ x = \frac{1}{e^{x}} \\ y = \frac{1}{e^{2}} \log (\frac{1}{e^{2}}) = - \frac{1}{e^{2}} \log e^{2} = - \frac{2}{e^{2}} \end{aligned}

\begin{aligned} Coordinates = [\frac{1}{e^{2}}, - \frac{2}{e^{2}}] \\ ∴ Equation= y + \frac{2}{e^{2}} = 1 (x - \frac{1}{e^{2}}) \\ y \cdot e^{2} + 2 = x e^{2} - 1 \\ e^{2} (x - y) = 3 \end{aligned}

Tangents and Normals Exercise 15.2 Question 13 Sub Question 1

Answer:
Equation of tangent

\Rightarrow y - 2 x - 3 = 0

Hint: Differentiate both with respect to x
Given:
parallel to

2 x - y + 9

Solution:
The equation of the given curve is

y = x^{2} - 2 x + 7

upon differentiation

\frac{d y}{d x} = 2 x - 2

This is in the form

y = m x + c

∴ s l o p e o f t h e l i n e = 2

If a tangent is parallel to the line

2 x - y + 9

= 0

then the slope of the tangent is equal to the slope of the line. Therefore we have:

\begin{aligned} 2 = 2 x - 2 \Rightarrow 2 x = 4 \Rightarrow x = 2 \\ Now, x = 2 \Rightarrow y = 4 - 4 + 7 = 7 \end{aligned}

Thus the equation of the tangent passing through (2,7) is given by

y - 7 = 2 (x - 2)

\Rightarrow y - 2 x - 3 = 0

Hence the equation of the tangent line to the given curve (which is parallel to the line

2 x - y + 9

= 0

) is

y - 2 x - 3 = 0

(Ans)

Tangents and Normals Exercise 15.2 Question 13 Sub Question 2

Answer:
Equation of tangent

\Rightarrow 36 y + 12 x - 227 = 0

Hint: Differentiate both with respect to x
Given:
perpendicular to

5 y - 15 x = 13

Solution:
the equation of the line is

5 y - 15 x = 13

5 y - 15 x = 13

∴ y = 3 x + \frac{13}{5}

This is of the form

y = m x + c

∴ t h e s l o p e o f t h e l i n e i s 3

If a tangent is perpendicular to the line

5 y - 15 x = 13

, then the slope of the tangent is

\begin{aligned} - \frac{1}{slope of the line} = - \frac{1}{3} \\ \Rightarrow 2 x - 2 = - \frac{1}{3} \\ \Rightarrow 2 x = - \frac{1}{3} + 2 \\ \Rightarrow 2 x = \frac{5}{3} \\ \Rightarrow x = \frac{5}{6} \end{aligned}

Now,

x = \frac{5}{6}

\Rightarrow y = \frac{25}{36} - \frac{10}{6} + 7 = \frac{217}{36}

Thus, the equation of the tangent passing through [5/6,217/36] is given by

\begin{aligned} \Rightarrow y - \frac{217}{36} = - \frac{1}{3} [x - \frac{5}{6}] \Rightarrow \frac{36 y - 217}{36} = - \frac{1}{18} (6 x - 5) \\ \Rightarrow 36 y - 217 = - 2 (6 x - 5) \Rightarrow 36 y - 217 = - 12 x + 10 \\ \Rightarrow 36 y + 12 x - 227 = 0 \\ the equation of the tangent is 36 y + 12 x - 227 = 0 \end{aligned}

(Ans)

Tangents and Normals Exercise 15.2 Question 14

Answer: $2 x^{2} - 12 x + 19 = 0$
Hint: find the slope of the curve
Given:

y = \frac{1}{x - 3}, x \neq 3

Solution: Slope of the line is 2
Slope of the curve is

\frac{-}{1} {(x - 3)}^{2}

\begin{aligned} \Rightarrow - \frac{1}{(x - 3)^{2}} = 2 \\ \Rightarrow 2 (x - 3)^{2} = - 1 \\ \Rightarrow 2 (x^{2} - 6 x + 9) = - 1 \\ \Rightarrow 2 x^{2} - 12 x + 19 = 0 \end{aligned}

Which has no real roots

b^{2} - 4 a c < 0

Hence: there is no tangent to the given curve having slope 2.

Tangents and Normals Exercise 15.2 Question 15

Answer:

y = \frac{1}{2}

Hint: Differentiate
Given:

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

Solution:

\begin{aligned} \frac{d y}{d x} = - \frac{1}{{(x^{2} - 2 x + 3)}^{2}} \times (2 x - 2) \\ ∴ - \frac{(2 x - 2)}{{(x^{2} - 2 x + 3)}^{2}} = 0 \Rightarrow x = 1 \\ when x = 1, y = \frac{1}{2} \\ ∴ Equation of the line y - \frac{1}{2} = 0 (x - 1) \\ y = \frac{1}{2} \Rightarrow 2 y = 1 or 2 y - 1 = 0 \end{aligned}

Tangents and Normals Exercise 15.2 Question 16

Answer: Equation of the tangent is

\Rightarrow 48 x - 24 y = 23

Hint:
Given:

y = \sqrt{3 x - 2}

Which is paralle to

y x - 2 y + 5 = 0

Solution: The equation of the given curve is

y = \sqrt{3 x - 2}

The slope of the tangent to the given curve at any point (x,y) is given by

\frac{d y}{d x} = \frac{2}{2 \sqrt{3 x - 2}}

The Equation of the given line is

4 x - 2 y + 5 = 0

4 x - 2 y + 5 = 0, ∴ y = 2 x + \frac{5}{2}

[w h i c h i s i n t h e f o r m o f y = m x + c]

∴ s l o p e o f l i n e i s 2

Now the tangent to the given curve is parallel to the line

4 x - 2 y - 5 = 0

, if slope of tangent is equal to the slope of line

\begin{aligned} \frac{3}{2 \sqrt{3 x - 2}} = 2 \Rightarrow \sqrt{3 x - 2} = \frac{3}{4} \\ \Rightarrow 3 x - 2 = \frac{9}{16} \Rightarrow 3 x = \frac{9}{16} + 2 = \frac{41}{16} \\ \Rightarrow x = \frac{41}{48} \\ When x = \frac{41}{48}, y = \sqrt{3 [\frac{41}{48}] - 2} = \frac{3}{4} \end{aligned}

the equation of tangent at point

[\frac{41}{48}, \frac{3}{4}]

is given by

\begin{aligned} y - \frac{3}{4} = 2 [x - \frac{41}{48}] \\ \Rightarrow \frac{4 y - 3}{4} = 2 [\frac{48 x - 41}{48}] \\ \Rightarrow 4 y - 3 = \frac{48 x - 41}{6} \\ \Rightarrow 48 x - 24 y = 23 \end{aligned}

Tangents and Normals Exercise 15.2 Question 17

Answer : Equation of the tangent is

4 x - y = - 13

Hint: Differentiate on both sides
Given:

x^{2} + 3 y = 3

which is parallel to

y - 4 x + 5 = 0

Solution: Given equation of the curve is

x^{2} + 3 y = 3

....(

i

)
On differentiating on both sides

2 x + 3 \frac{d y}{d x} = 0

\frac{d y}{d x} = - \frac{2 x}{3}

∴

the slope of tangent

(m) = - \frac{2 x}{3}

Given the equation of the line

$y - 4 x + 5 = 0$ or $y = y x - 5$

$w h i c h i s o f t h e f o r m y = m x + c$

$∴ s l o p e o f t a n g e n t = s l o p e o f l i n e$

or $- \frac{2 x}{3} = 4$ or $- 2 x = 12$

$\Rightarrow x = - 6$

on putting $\Rightarrow x = - 6$ in the Eq $(i)$ we get

$\begin{aligned} (- 6)^{2} + 3 y = 3 or 3 y = 3 - 36 \\ or 3 y = - 33 or y = - 11 \end{aligned}$

so,the tanent is passing through point $(- 6, 11)$ and it has slope 4
Hence the required equation of the tangent is

$\begin{aligned} y + 11 = 4 (x + 6) \\ y + 11 = 4 x + 24 \\ 4 x - y = - 13 \end{aligned}$ (Ans)

Tangents and Normals Exercise 15.2 Question 18

Answer: To Prove

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

(R.H.S=L.H.S)
Hint: Differentiate
Given:

{[\frac{x}{a}]}^{n} + {[\frac{y}{b}]}^{n} = 2

Solution:

{[\frac{x}{a}]}^{n} + {[\frac{y}{b}]}^{n} = 2

\begin{aligned} \Rightarrow \frac{n}{a} {[\frac{x}{a}]}^{n - 1} + \frac{n}{b} {[\frac{y}{b}]}^{n - 1} \frac{d y}{d x} = 0 \\ \Rightarrow \frac{n}{b} {[\frac{y}{b}]}^{n - 1} \frac{d y}{d x} = - \frac{n}{a} {[\frac{x}{a}]}^{n - 1} \\ \Rightarrow \frac{d y}{d x} = - \frac{n}{a} {[\frac{x}{a}]}^{n - 1} \times \frac{b}{n} {[\frac{b}{y}]}^{n - 1} = - \frac{b}{a} {(\frac{b x}{a y})}^{n - 1} \end{aligned}

∴ s l o p e o f t a n g e n t =

{(\frac{d y}{d x})}_{(a, b)} = - \frac{b}{a} {(\frac{b \times a}{a \times b})}^{n - 1} = - \frac{b}{a}

....(ii)
The equation of the tangent is

$\begin{aligned} y - b = - \frac{b}{a} (x - a) \\ \Rightarrow y a - a b = - x b + a b \\ \Rightarrow x b + y a = 2 a b \\ \Rightarrow \frac{x}{a} + \frac{y}{b} = 2 \end{aligned}$
So the given line touched the given curve at the given point

Tangents and Normals Exercise 15.2 Question 19

Answer:

3 y = 2 \sqrt{2 x} - 2

Hint: Differentiate to find the slop of the tangent
Given:

x = \sin 3 t, y = \cos t 2 t

t = \frac{π}{4}

Solution: Slope of tangent is

\frac{d y}{d x}

\begin{aligned} \Rightarrow \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{\frac{d (\cos 2 t)}{d t}}{\frac{d (\sin 3 t)}{d t}} = \frac{- 2 \sin 2 t}{3 \cos 3 t} \\ {[\frac{d y}{d x}]}_{x t - \frac{π}{4}} = \frac{- 2 \sin \frac{π}{2}}{3 \cos \frac{3 π}{4}} = \frac{- 2 \times 1}{3 \times [- \frac{1}{\sqrt{2}}]} = \frac{2 \sqrt{2}}{3} \end{aligned}

Now,

x = \sin [\frac{3 π}{4}] = \frac{1}{\sqrt{2}}

y = \cos [\frac{2 π}{4}] = 0

$∴$ Equation of the tangent is

$\begin{aligned} y - 0 = \frac{d y}{d x} [x - [\frac{1}{\sqrt{2}}]] \\ y = \frac{2 \sqrt{2}}{3} [x - \frac{1}{\sqrt{2}}] \\ y = \frac{2 \sqrt{2}}{3} x - \frac{2}{3} \end{aligned}$

or $3 y = 2 \sqrt{2 x} - 2$

Tangents and Normals Exercise 15.2 Question 20

Answer:

y - 6 = 0

and y =7 at points (2,7),(3,6)
Hint: Differentiate both side with respect to x
Given :

y = 2 x^{3} - 15 x^{2} + 36 x - 21

Differentiating both sides w.r.t x, we get

\frac{d y}{d x} = 6 x^{2} - 30 x + 36

For the points on the curve, where tangents are parallel to x-axis

\frac{d y}{d x} = 0

\begin{aligned} \Rightarrow 6 x^{2} - 30 x + 36 = 0 \Rightarrow x^{2} - 5 x + 6 = 0 \\ ∴ (x - 2) (x - 3) = 0 \Rightarrow x = 2, 3 \\ ∴ from (1), y = 2 (2)^{3} - 15 (2)^{2} + 36 (2) - 21, 2 (3)^{2} - 15 (3)^{2} + 36 (3) - 21 \\ = 16 - 60 + 72 - 21, 54 - 135 + 108 - 21 = 7, 6 \\ ∴ points are (2, 7), (3, 6) \end{aligned}

The equation of tangent at (2,7) parallel to x-axis is

y - 7 = 0 (x = 2)

[∵ m = 0]

The equation of tangent at (2,6) parallel to x-axis is

y - 6 = 0 (x = 3)

y - 6 = 0

Tangents and Normals Exercise 15.2 Question 21

Answer: The equation of tangent

\to 3 x + y = 4, y = 3 x - 4

Hint: Differentiate with respect to

x

.
Given:

3 x^{2} - y^{2} = 8

, which passes through

(\frac{4}{3}, 0)

Solution:

3 x^{2} - y^{2} = 8

. . . . (i)

Diff w.r.t

x

\begin{aligned} 6 x - 2 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = \frac{3 x}{y} \\ \Rightarrow {[\frac{d y}{d x}]}_{(x_{i}, y_{i})} = \frac{3 x_{1}}{y_{1}} \end{aligned}

$∴$ The eq. of tangent at $(x_{1}, y_{1})$ is

$y - y_{1} = \frac{3 x_{1}}{y_{1}} (x - x_{1})$

Tangent passes through the point _{$[\frac{4}{3}, 0]$}

_{$\begin{aligned} ∴ 0 - y_{1} = \frac{3 x_{1}}{y_{1}} [\frac{4}{3} - x_{1}] \\ \begin{matrix} - y_{1}^{2} = 3 x_{1} [\frac{4}{3} - x_{1}] \Rightarrow - y_{1}^{2} = 4 x_{1} - 3 x_{1}^{2} \\ Or \end{matrix} \\ Using egn (i) 8 - 3 x_{1}^{2} = 4 x_{1} - 3 x_{1}^{2} \end{aligned}$}

_{$o r x_{1} = 2$}

_{$\begin{aligned} ∴ 3 (2^{2}) - y_{1}^{2} = 8 \\ 12 - y_{1}^{2} = 8 \Rightarrow y_{1}^{2} = 4 \\ y_{1}^{2} = \pm 2 \end{aligned}$}

$∴$ the points are (2,2) and (2,-2)

$∴$ The eq. of tangent at (2,2) is :

$y - 2 = 3 (x - 2)$

$y = 3 x - 4$

The eq.of tangent at (2,-2) is :

$y + 2 = - 3 (x - 2)$

$y = - 3 x + 4$

$3 x + y = 4$

Tangents and Normals Exercise 15.2 Question 3 Sub Question 19

Answer: Equation of tangent, $\sqrt{2} b x - a y - a b = 0$
Equation of normal, $a x + \sqrt{2} b y - \sqrt{2} (a^{2} + b^{2}) = 0$
Hint: Differentiating the given curve with respect to x.
Given: $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 a t (\sqrt{2 a}, b)$
Solution: $\frac{d y}{d x} = \frac{b^{2} x}{a^{2} y}$
${(\frac{d y}{d x})}_{\sqrt{2 a, b}} = \frac{\sqrt{2 b}}{a}$
Slope of tangent is
$\frac{\sqrt{2 b}}{a}$ and of the normal is
The equation of tangent is
$\begin{aligned} y - b = \frac{\sqrt{2 b}}{a} (x - \sqrt{2} a) \\ \sqrt{2} b x - a y - a b = 0 \end{aligned}$
Equation of normal is,
$\begin{aligned} y - b = \frac{- a}{\sqrt{2} b} (x - \sqrt{2} a) \\ a x + \sqrt{2} b y - \sqrt{2} (a^{2} + b^{2}) = 0 \end{aligned}$

The 15th chapter, Tangents, and Normals are not where students find the sums very easy to solve. Predominantly, the second exercise, ex 15.2, consists of sums that are tricky to solve. It includes concepts like finding the equation of the tangent, finding the normal of the tangents to specific points, equation of the normal to the curve, and equation of lines to the slope. There are around 45 questions in this exercise, including its subparts. Again, the RD Sharma Class 12 Chapter 15 Exercise 15.2 will lend a helping hand.

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