RD Sharma Class 12 Exercise 15.1 Tangents and Normals Solutions Maths

RD Sharma Class 12 Exercise 15.1 Tangents and Normals Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 21, 2022 01:10 PM IST

Students studying in the CBSE board schools possess the set of RD Sharma solution books for their reference. When they are unable to clarify their doubts with the teacher, these books will help them out. Mathematics is a challenging subject for every student. The Class 12 students are no exception to it.RD Sharma solution Significantly, in the 15th chapter, Tangents and Normals tend to be more challenging. In such cases, the RD Sharma Class 12th Exercise 15.1 books lend a helping hand.

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

Chapter 15 - Tangents and Normals Ex 15.2

Chapter 15 - Tangents and Normals Ex 15.3

Chapter 15 -Tangents and Normals Ex-MCQ

Chapter 15 -Tangents and Normals Ex-FBQ

Chapter 15 -Tangents and Normals Ex-VSA

Tangents and Normals Excercise: 15.1

Tangents and Normals Exercise 15.1 Question 1 Sub Question 1 .

Answer: $T h e S l o p e o f t h e t a n g e n t i s 3$
$T h e S l o p e o f t h e n o r m a l s i s \frac{1}{3}$
Hint:

Tangents and Normals Exercise 15.1 Question 1 Sub Question 2 .

Answer:
$T h e s l o p e o f t h e t a n g e n t i s \frac{1}{6}$
$The slope of the normal is - 6$
Hint:
$The slope of tangent is the limit of \frac{Δ y}{Δ x} as Δ x approaches zero.$
$The slope of normal is the normal to a curve at P = (x, y) is a line perpendicular to the tangent$
$at P and passing through P .$
Given: $y = \sqrt{x} at x = 9$
Solution:
$First we have to find \frac{d y}{d x} of given function,$
$f (x) that is to find the derivative of f (x)$
$y = \sqrt{x}$
$\begin{aligned} ∴ \sqrt[n]{x} = x^{\frac{1}{n}} \\ y = (x)^{\frac{1}{2}} \\ \frac{d}{d x} (x^{n}) = n x^{n - 1} \end{aligned}$
$We know that the slope of the tangent is \frac{d y}{d x}$
$\begin{aligned} \frac{d y}{d x} \Rightarrow \frac{1}{2} (x)^{\frac{1}{2} - 1} \\ \frac{d y}{d x} \Rightarrow \frac{1}{2} (x)^{- \frac{1}{2}} \end{aligned}$
Since, $x = 9$
$\begin{aligned} {(\frac{d y}{d x})}_{x - 9} \Rightarrow \frac{1}{2} (9)^{\frac{- 1}{2}} \\ {(\frac{d y}{d x})}_{x - 9} \Rightarrow \frac{1}{2} \times \frac{1}{\sqrt{9}} \\ {(\frac{d y}{d x})}_{x - 9} \Rightarrow \frac{1}{2} \times \frac{1}{3} \\ {(\frac{d y}{d x})}_{x - 9} \Rightarrow \frac{1}{6} \end{aligned}$
$The slope of the tangent at x = is \frac{1}{6}$
$The slope of the normal = \frac{- 1}{The slope of the tangent}$
$The slope of the normal = \frac{- 1}{{(\frac{d y}{d x})}_{x - 9}}$
$The slope of the normal = \frac{- 1}{\frac{1}{6}} = - 6$

Tangents and Normals Exercise 15.1 Question 1 Sub Question 3 .

Answer:
$The slope of the tangent is 11$
$The slope of the normal is \frac{- 1}{11}$
Hint:
$The slope of tangent is the limit of \frac{Δ y}{Δ x} as Δ x approaches zero.$
$The slope of normal is the normal to a curve at P = (x, y) is a line perpendicular to the tangent$
$at P and passing through P .$
Given: $y = x^{3} - x at x = 2$
Solution:
$First we have to find \frac{d y}{d x} of given function,$
$f (x) that is to find the derivative of f (x)$
$y = x^{3} - x$
$\frac{d}{d x} (x^{n}) = n x^{n - 1}$
$We know that the slope of the tangent is \frac{d y}{d x}$
$\begin{aligned} \frac{d y}{d x} = 3 x^{3 - 1} - 1 x^{1 - 1} \\ \frac{d y}{d x} = 3 x^{2} - 1 {x^{0} = 1} \end{aligned}$
Since,

x = 2

\begin{aligned} {(\frac{d y}{d x})}_{x = 2} \Rightarrow 3 (2)^{2} - 1 \\ {(\frac{d y}{d x})}_{x - 2} \Rightarrow 3 (4) - 1 \\ {(\frac{d y}{d x})}_{x - 2} \Rightarrow 12 - 1 \\ {(\frac{d y}{d x})}_{x - 2} \Rightarrow 11 \end{aligned}

The slope of the tangent at x = 2 is 11

The slope of the normal = \frac{- 1}{The slope of the tangent}

The slope of the normal = \frac{- 1}{{(\frac{d y}{d x})}_{x - 2}}

The slope of the normal = \frac{- 1}{11}

Tangents and Normals Exercise 15.1 Question 1 Sub Question 4

Answer:

The slope of the tangent is 3

The slope of the normal is \frac{- 1}{3}

Hint

The slope of tangent is the limit of \frac{Δ y}{Δ x} as Δ x approaches zero.

The slope of normal is the normal to a curve at P = (x, y) is a line perpendicular to the tangent

at P and passing through P .

Given:

y = 2 x^{2} + 3 \sin x at x = 0

Solution:

First we have to find \frac{d y}{d x} of given function,

f (x) that is to find the derivative of f (x)

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

\begin{aligned} \frac{d}{d x} (x^{n}) = n x^{n - 1} \\ \frac{d}{d x} (\sin x) = \cos x \end{aligned}

We know that the slope of the tangent is \frac{d y}{d x}

\begin{aligned} \frac{d y}{d x} = 2 (2 x^{2 - 1}) + 3 \cos x \\ \frac{d y}{d x} = 4 x + 3 \cos x \end{aligned}

Since,

x = 0

{(\frac{d y}{d x})}_{x = 0} \Rightarrow 4 (0) - 3 \cos (0)

{(\frac{d y}{d x})}_{x = 0} \Rightarrow 0 + 3 (1) {\cos (0) = 1}

{(\frac{d y}{d x})}_{x - 0} \Rightarrow 3

The slope of the tangent at x = 0 is 3

The slope of the normal = \frac{- 1}{The slope of the tangent}

The slope of the normal = \frac{- 1}{{(\frac{d y}{d x})}_{x - 0}}

The slope of the normal = \frac{- 1}{3}

Tangents and Normals Exercise 15.1 Question 1 Sub Question 5

Answer:
$The slope of the tangent is 1$
$The slope of the normal is - 1$
Hint:
$The slope of tangent is the limit of \frac{Δ y}{Δ x} as Δ x approaches zero.$
$The slope of normal is the normal to a curve at P = (x, y) is a line perpendicular to the tangent$ $at P and passing through P .$
Given: $x = a (θ - \sin θ), y = a (1 + \cos θ) at θ = \frac{- π}{2}$
$Here to find \frac{d y}{d x}, we have to find \frac{d y}{d θ} & \frac{d x}{d θ} and$
$divide \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} and we get desired \frac{d y}{d x}$
Solution:
$\begin{aligned} \frac{d}{d x} (x^{n}) = n x^{n - 1} \\ \frac{d}{d x} (\sin x) = \cos x \\ x = a (θ - \sin θ) \\ \frac{d x}{d θ} = a {\frac{d}{d θ} (θ) - \frac{d}{d θ} (\sin θ)} \end{aligned}$
$\frac{d x}{d θ} = a (1 - \cos θ) \to (1)$
$y = a (1 + \cos θ)$
$\frac{d y}{d θ} = a {\frac{d}{d θ} (1) + \frac{d}{d θ} (\cos θ)}$
$\frac{d}{d x} (constant) = 0$
$\frac{d}{d x} (\cos x) = - \sin x$
$\frac{d y}{d θ} = a (0 + (- \sin θ))$
$\frac{d y}{d θ} = a (- \sin θ)$
$\frac{d y}{d θ} = - a \sin θ \to (2)$
$\frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} \Rightarrow \frac{- a \sin θ}{a (1 - \cos θ)}$
$\frac{d y}{d x} = \frac{- \sin θ}{(1 - \cos θ)}$
$The slope of the tangent is \frac{- \sin θ}{(1 - \cos θ)}$
$Since, θ = \frac{- π}{2}$
${(\frac{d y}{d x})}_{θ - \frac{- π}{2}} \Rightarrow \frac{- \sin (\frac{- π}{2})}{(1 - \cos \frac{- π}{2})}$
$We know that \cos (\frac{- π}{2}) = 0 and \sin (\frac{- π}{2}) = 1$
$\begin{aligned} {(\frac{d y}{d x})}_{θ_{- \frac{- π}{2}}} \Rightarrow \frac{- (- 1)}{(1 - (- 0))} \\ {(\frac{d y}{d x})}_{θ_{- \frac{- π}{2}}} \Rightarrow \frac{1}{1 - 0} \Rightarrow 1 \end{aligned}$
$The slope of the tangent at θ = \frac{- π}{2} is 1$
$The slope of the normal = \frac{- 1}{The slope of the tangent}$
$The slope of the normal = \frac{- 1}{{(\frac{d y}{d x})}_{θ - \frac{- π}{2}}}$
$The slope of the normal = \frac{- 1}{1} = - 1$

Tangents and Normals Exercise 15.1 Question 1 Sub Question 6

Answer:
$The slope of the tangent is - 1$
$The slope of the normal is 1$
Hint
$The slope of tangent is the limit of \frac{Δ y}{Δ x} as Δ x approaches zero.$
$The slope of normal is the normal to a curve at P = (x, y) is a line perpendicular to the tangent$
$at P and passing through P .$
Given: $x = a \cos^{3} θ, y = a \sin^{3} θ at θ = \frac{π}{4}$
Solution:
$Here to find \frac{d y}{d x}, we have to find \frac{d y}{d θ} & \frac{d x}{d θ} and$
$divide \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} and we get desired \frac{d y}{d x}$
$\begin{aligned} \frac{d}{d x} (x^{n}) = n x^{n - 1} \\ \frac{d}{d x} (\cos x) = - \sin x \\ x = a \cos^{3} θ \\ \frac{d x}{d θ} = a {\frac{d}{d θ} (\cos^{3} θ)} \end{aligned}$
$\frac{d x}{d θ} = a (3 \cos^{3 - 1} θ \times - \sin θ)$
$\frac{d x}{d θ} = - 3 a \cos^{2} θ \sin θ \to (1)$
$y = a \sin^{3} θ$
$\frac{d y}{d θ} = a {\frac{d}{d θ} (\sin^{3} θ)}$
$\frac{d}{d x} (\sin x) = \cos x$
$\frac{d y}{d θ} = a (3 \sin^{3 - 1} θ \times \cos θ)$
$\frac{d y}{d θ} = 3 a \sin^{2} θ \cos θ \to (2)$
$\frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} \Rightarrow \frac{3 a \sin^{2} θ \cos θ}{- 3 a \cos^{2} θ \sin θ}$
$\frac{d y}{d x} = \frac{\sin θ}{- \cos θ}$
$\frac{d y}{d x} = - \tan θ$
$The slope of the tangent is - \tan θ$
$Since, θ = \frac{π}{4}$
${(\frac{d y}{d x})}_{θ = \frac{π}{4}} \Rightarrow - \tan \frac{π}{4} = - 1$
$The slope of the tangent at θ = \frac{π}{4} is - 1$
$The slope of the normal = \frac{- 1}{The slope of the tangent}$
$The slope of the normal = \frac{- 1}{{(\frac{d y}{d x})}_{θ = \frac{- π}{2}}}$
$The slope of the normal = \frac{- 1}{- 1} = 1$

Tangents and Normals Exercise 15.1 Question 1 Sub Question 7

Answer:
$The slope of the tangent is 1$
$The slope of the normal is - 1$
Hint:
$The slope of tangent is the limit of \frac{Δ y}{Δ x} as Δ x approaches zero.$
$The slope of normal is the normal to a curve at P = (x, y) is a line perpendicular to the tangent$
$at P and passing through P .$
Given: $x = a (θ - \sin θ) y = a (1 - \cos θ) at θ = \frac{π}{2}$
Solution:
$Here to find \frac{d y}{d x}, we have to find \frac{d y}{d θ} & \frac{d x}{d θ} and$
$divide \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} and we get desired \frac{d y}{d x}$
$\begin{aligned} \frac{d}{d x} (x^{n}) = n x^{n - 1} \\ \frac{d}{d x} (\sin x) = \cos x \end{aligned}$
$\begin{aligned} x = a (θ - \sin θ) \\ \frac{d x}{d θ} = a {\frac{d x}{d θ} (θ) - \frac{d x}{d θ} (\sin θ)} \\ \frac{d x}{d θ} = a (1 - \cos θ) \to (1) \end{aligned}$
$y = a (1 - \cos θ)$
$\frac{d y}{d θ} = a {\frac{d y}{d θ} (1) - \frac{d y}{d θ} (\cos θ)}$
$\frac{d}{d x} (constant) = 0$
$\frac{d}{d x} (\cos x) = - \sin x$
$\frac{d y}{d θ} = a (0 - (- \sin θ))$
$\frac{d y}{d θ} = a \sin θ \to (2)$
$\frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} \Rightarrow \frac{a \sin θ}{a (1 - \cos θ)}$
$\frac{d y}{d x} = \frac{\sin θ}{(1 - \cos θ)}$
$Since, θ = \frac{π}{2}$
${(\frac{d y}{d x})}_{θ = \frac{π}{2}} = \frac{\sin \frac{π}{2}}{(1 - \cos \frac{π}{2})}$
$We know that \cos (\frac{π}{2}) = 0 and \sin (\frac{π}{2}) = 1$
${(\frac{d y}{d x})}_{θ = \frac{π}{2}} = \frac{1}{(1 - 0)} = 1$
$The slope of the tangent at θ = \frac{π}{2} is 1$
$The slope of the normal = \frac{- 1}{The slope of the tangent}$
$The slope of the normal = \frac{- 1}{{(\frac{d y}{d x})}_{θ = \frac{- π}{2}}}$
$The slope of the normal = \frac{- 1}{1} = - 1$

Tangents and Normals Exercise 15.1 Question 1 Sub Question 8 .

Answer:
$The slope of the tangent is - 12$
$The slope of the normal is \frac{1}{12}$
Hint:
$The slope of tangent is the limit of \frac{Δ y}{Δ x} as Δ x approaches zero.$
$The slope of normal is the normal to a curve at P = (x, y) is a line perpendicular to the tangent$
$at P and passing through P .$
Given: $y = (\sin 2 x + \cot x + 2)^{2} at x = \frac{π}{2}$
Solution:
$First we have to find \frac{d y}{d x} of given function,$
$f (x) that is to find the derivative of f (x)$
$\begin{aligned} \frac{d}{d x} (x^{n}) = n x^{n - 1} \\ y = (\sin 2 x + \cot x + 2)^{2} \end{aligned}$
$\frac{d y}{d x} = 2 (\sin 2 x + \cot x + 2)^{2 - 1} + {\frac{d}{d x} (\sin 2 x) + \frac{d}{d x} (\cot x) + \frac{d}{d x} (2)}$
$\frac{d y}{d x} = 2 (\sin 2 x + \cot x + 2) {(\cos 2 x) 2 + (- \cos e c^{2} x) + (0)}$
$\begin{aligned} \frac{d}{d x} (\sin x) = \cos x \\ \frac{d}{d x} (\cot x) = - {cosec}^{2} x \\ \frac{d y}{d x} = 2 (\sin 2 x + \cot x + 2) (2 \cos 2 x - \cos e c^{2} x) \end{aligned}$
$Since, x = \frac{π}{2}$
${(\frac{d y}{d x})}_{x = \frac{π}{2}} = 2 (\sin 2 (\frac{π}{2}) + \cot (\frac{π}{2}) + 2) (2 \cos 2 (\frac{π}{2}) - {cosec}^{2} (\frac{π}{2}))$
${(\frac{d y}{d x})}_{x = \frac{π}{2}} = 2 \times (0 + 0 + 2) (2 (- 1) - 1)$
$We know that \cot (\frac{π}{2}) = 0 and cosec (\frac{π}{2}) = 1$
${(\frac{d y}{d x})}_{x = \frac{π}{2}} = 2 \times (2) (- 2 - 1)$
${(\frac{d y}{d x})}_{x = \frac{π}{2}} = 4 \times (- 3) = - 12$
$The slope of the tangent at x = \frac{π}{2} is - 12$
$The slope of the normal = \frac{- 1}{The slope of the tangent}$
$The slope of the normal = \frac{- 1}{{(\frac{d y}{d x})}_{x = \frac{π}{2}}}$
$The slope of the normal = \frac{1}{12}$

Tangents and Normals Exercise 15.1 Question 1 Sub Question 9

Answer:
$The slope of the tangent is - \frac{2}{5}$
$The slope of the normal is \frac{5}{2}$
Hint:
$The slope of tangent is the limit of \frac{Δ y}{Δ x} as Δ x approaches zero.$
$The slope of normal is the normal to a curve at P = (x, y) is a line perpendicular to the tangent$
$at P and passing through P .$
Given: $x^{2} + 3 y + y^{2} = 5 at (1, 1)$
$Here we have to differentiate the equation with respect to x$
Solution:
$\frac{d}{d x} (x^{2} + 3 y + y^{2}) = \frac{d}{d x} (5)$
$\frac{d}{d x} (x^{2}) + \frac{d}{d x} (3 y) + \frac{d}{d x} (y^{2}) = \frac{d}{d x} (5)$
$\frac{d}{d x} (x^{n}) = n x^{n - 1}$
$\Rightarrow 2 x + 3 \frac{d y}{d x} + 2 y \frac{d y}{d x} = 0$
$\Rightarrow 2 x + \frac{d y}{d x} (3 + 2 y) = 0$
$\Rightarrow \frac{d y}{d x} (3 + 2 y) = - 2 x$
$\frac{d y}{d x} = \frac{- 2 x}{(3 + 2 y)}$
$The slope of tangent at (1, 1) is$
$\frac{d y}{d x} = \frac{- 2 \times 1}{(3 + 2 \times 1)} = \frac{- 2}{3 + 2}$
$\frac{d y}{d x} = \frac{- 2}{5}$
$The slope of the tangent at (1, 1) is \frac{- 2}{5}$
$The slope of the normal = \frac{- 1}{The slope of the tangent}$
$The slope of the normal = \frac{- 1}{(\frac{d y}{d x})}$
$The slope of the normal = \frac{- 1}{\frac{- 2}{5}} = \frac{5}{2}$

Tangents and Normals Exercise 15.1 Question 1 Sub Question 10 .

Answer:
$The slope of the tangent is - 6$
$The slope of the normal is \frac{1}{6}$
Hint:
$The slope of tangent is the limit of \frac{Δ y}{Δ x} as Δ x approaches zero.$
$The slope of normal is the normal to a curve at P = (x, y) is a line perpendicular to the tangent$
$at P and passing through P .$
Given: $x y = 6 at (1, 6)$

Here we have to use product rule,

Solution:

\frac{d}{d x} (x y) = \frac{d}{d x} (6)

x \frac{d}{d x} (y) + y \frac{d}{d x} (x) = \frac{d}{d x} (6)

\frac{d}{d x} (constant) = 0

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

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

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

The slope of tangent at (1, 6) is

\frac{d y}{d x} = \frac{- 6}{1} = - 6

The slope of the tangent at (1, 6) is - 6

The slope of the normal = \frac{- 1}{The slope of the tangent}

The slope of the normal = \frac{- 1}{(\frac{d y}{d x})}

The slope of the normal = \frac{- 1}{- 6} = \frac{1}{6}

Tangents and Normals Exercise 15.1 Question 2 .

Answer: $a = 5 and b = - 4$
Hint: $The slope of the tangent is the limit of \frac{Δ y}{Δ x} as Δ x approaches zero.$
Given:
$The slope of the tangent to the curve x y + a x + b y = 2 at (1, 1) is 2$
Solution:
$First we will find the slope of tangent by using product rule, we get$
$x y + a x + b y = 2$
$x \frac{d}{d x} (y) + y \frac{d}{d x} (x) + a \frac{d}{d x} (x) + b \frac{d}{d x} (y) = \frac{d}{d x} (2)$
$x \frac{d y}{d x} + y + a + b \frac{d y}{d x} = 0$
$\frac{d y}{d x} (x + b) = - (a + y)$
$\frac{d y}{d x} = \frac{- (a + y)}{(x + b)}$
$The slope of tangent to the curve x y + a x + b y = 2 at (1, 1) is 2$
$\frac{d y}{d x} = 2$
$\Rightarrow \frac{- (a + y)}{(x + b)} = 2$
$\Rightarrow \frac{- (a + 1)}{(1 + b)} = 2$
$\Rightarrow - (a + 1) = 2 (1 + b)$
$\Rightarrow - (a + 1) = 2 + 2 b$
$\Rightarrow a + 2 b = - 3 \to (1)$
$Also the point (1, 1) lies on the curve x y + a x + b y = 2 we have,$
$\begin{aligned} 1 \times 1 + a \times 1 + b \times 1 = 2 \\ 1 + a + b = 2 \\ a + b = 1 \to (2) \end{aligned}$
$\begin{aligned} Subtract eqn(1) and (2) \\ a + 2 b = - 3 \\ \frac{a + b = 1}{b = - 4} \end{aligned}$
$Substitute b = - 4 in a + b = 1$
$\begin{aligned} a - 4 = 1 \\ a = 5 \\ So, a = 5 and b = - 4 \end{aligned}$

Tangents and Normals Exercise 15.1 Question 3 .

Answer: $a = - 2 and b = - 5$
Hint: $The slope of the tangent is the limit of \frac{Δ y}{Δ x} as Δ x approaches zero.$
Given:
$The slope of the tangent to the curve y = x^{3} + a x + b at (1, - 6)$
Solution:
First we will find the slope of tangent, we get

y = x^{3} + a x + b

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

\frac{d y}{d x} = 3 x^{3 - 1} + a \frac{d x}{d x} + 0

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

The slope of tangent to the curve y = x^{3} + a x + b at (1, - 6) is

\frac{d y}{d x (x = 1, y = - 6)} = 3 (1)^{2} + a

\Rightarrow 3 + a \to (1)

The given line is x - y + 5 = 0

y = x + 5 is the form of equation of a straight line y = m x + c

where m is the slope of line.

So, the slope of the line is y = 1 \times x + c

so, the slope is 1 \to (2)

Also the point (1, - 6) lie on the tangent

x = 1 & y = - 6 satisfies the equation

y = x^{3} + a x + b

- 6 = (1)^{3} + a \times 1 + b

- 6 = 1 + a + b

a + b = - 7 \to (3)

Since the tangent is parallel to the line, from (1) \& (2)

3 + a = 1

a = - 2

from (3)

a + b = - 7

- 2 + b = - 7

b = - 7 + 2

b = - 5

so, the value is a = - 2 and b = - 5

Tangents and Normals Exercise 15.1 Question 4

Answer: $x = \pm \sqrt{\frac{7}{3}} & y = \pm \frac{- 2}{3} \sqrt{\frac{7}{3}}$
Hint: $The slope of the tangent is the limit of \frac{Δ y}{Δ x} as Δ x approaches zero.$
Given: $Curve y = x^{3} - 3 x$
Solution:
$First we will find the slope of tangent, we get$
$y = x^{3} - 3 x$
$\frac{d y}{d x} = \frac{d}{d x} (x^{3}) - \frac{d}{d x} (3 x)$
$\frac{d y}{d x} = 3 x^{3 - 1} - 3 \frac{d x}{d x}$
$\frac{d y}{d x} = 3 x^{2} - 3 \to (1)$
$The equation of line passing through (x_{0}, y_{0}) and the slope m is y - y_{0} = m (x - x_{0})$
$So the slope m = \frac{y - y_{0}}{x - x_{0}}$
$The slope of the chord joining (1, - 2) & (2, 2)$
$\frac{d y}{d x} = \frac{2 - (- 2)}{2 - 1} = \frac{4}{1}$
$\frac{d y}{d x} = 4 \to (2)$
$from (1) \&(2)$
$3 x^{2} - 3 = 4$
$3 x^{2} = 4 + 3$
$3 x^{2} = 7$
$x^{2} = \frac{7}{3}$
$x = \pm \sqrt{\frac{7}{3}}$
$y = x^{3} - 3 x$
$y = x (x^{2} - 3)$
$y = \pm \sqrt{\frac{7}{3}} ({(\pm \sqrt{\frac{7}{3}})}^{2} - 3)$
$y = \pm \sqrt{\frac{7}{3}} (\frac{7}{3} - 3)$
$y = \pm \sqrt{\frac{7}{3}} (\frac{7 - 9}{3})$
$y = \pm \sqrt{\frac{7}{3}} (\frac{- 2}{3})$
$y = \pm (\frac{- 2}{3}) \sqrt{\frac{7}{3}}$
$Thus, the required point is x = \pm \sqrt{\frac{7}{3}} & y = \pm \frac{- 2}{3} \sqrt{\frac{7}{3}}$

Tangents and Normals Exercise 15.1 Question 5

Answer:

The points are (2, - 4) & (\frac{- 2}{3}, \frac{4}{27})

Hint:

The slope of the tangent is the limit of \frac{Δ y}{Δ x} as Δ x approaches zero.

Given:

Curve y = x^{3} - 2 x^{2} - 2 x and a line y = 2 x - 3

Solution:
First we will find the slope of tangent, we get

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

\begin{aligned} \frac{d}{d x} (x^{n}) = n x^{n - 1} \\ \frac{d y}{d x} = \frac{d}{d x} (x^{3}) - \frac{d}{d x} (2 x^{2}) - \frac{d}{d x} (2 x) \\ \frac{d y}{d x} = 3 x^{3 - 1} - 2 \times 2 (x^{2 - 1}) - 2 \times x^{1 - 1} \\ \frac{d y}{d x} = 3 x^{2} - 4 x - 2 \to (1) \end{aligned}

y = 2 x - 3 is the form of equation of a straight line y = m x + c

where m is the slope of line.

So, the slope of the line is y = 2 \times x - 3

so, the slope is 2 \to (2)

from (1) \& (2)

\begin{aligned} 3 x^{2} - 4 x - 2 = 2 \\ 3 x^{2} - 4 x = 4 \\ 3 x^{2} - 4 x - 4 = 0 \end{aligned}

We will use factorization method to solve the above quadratic equation.

\begin{aligned} 3 x^{2} - 6 x + 2 x - 4 = 0 \\ 3 x (x - 2) + 2 (x - 2) = 0 \\ (x - 2) (3 x + 2) = 0 \end{aligned}

x - 2 = 0 & 3 x + 2 = 0

x = 2 or x = \frac{- 2}{3}

Substitute x = 2 & x = \frac{- 2}{3} in y = x^{3} - 2 x^{2} - 2 x

When x = 2

\begin{aligned} y = (2)^{3} - 2 \times (2)^{2} - 2 \times (2) \\ y = 8 - 2 \times 4 - 4 \\ y = 8 - 8 - 4 \\ y = - 4 \end{aligned}

When x = \frac{- 2}{3}

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

y = \frac{- 8}{27} - 2 \times \frac{4}{9} + \frac{4}{3}

y = \frac{- 8}{27} - 2 \times \frac{4}{9} + \frac{4}{3}

y = \frac{- 8}{27} - \frac{8}{9} + \frac{4}{3}

L.C.M of 27, 3 and 9 is 27

y = \frac{(- 8 \times 1) - (8 \times 3) + (4 \times 9)}{27}

y = \frac{- 8 - 24 + 36}{27} = \frac{4}{27}

Thus, the points are (2, - 4) & (\frac{- 2}{3}, \frac{4}{27})

Tangents and Normals Exercise 15.1 Question 6

Answer: $The required point is (2, 4)$
Hint: $The slope of the tangent is the limit of \frac{Δ y}{Δ x} as Δ x approaches zero.$
Given: $The curve y^{2} = 2 x^{3} and the slope of tangent is 3$
Solution: $y^{2} = 2 x^{3}$
$\begin{aligned} Differentiating the above with respect to x \\ ∴ \frac{d}{d x} (x^{n}) = n x^{n - 1} \\ 2 y^{2 - 1} \frac{d y}{d x} = 2 \times (3 x)^{3 - 1} \end{aligned}$
$\begin{aligned} 2 y \frac{d y}{d x} = 2 \times 3 x^{2} \\ y \frac{d y}{d x} = 3 x^{2} \\ \frac{d y}{d x} = \frac{3 x^{2}}{y} \end{aligned}$
Since the slope of tangent is 3

\begin{aligned} \frac{3 x^{2}}{y} = 3 \\ \frac{x^{2}}{y} = 1 \\ x^{2} = y \end{aligned}

Substituting x^{2} = y in y^{2} = 2 x^{3}

{(x^{2})}^{2} = 2 x^{3}

x^{4} - 2 x^{3} = 0

x^{3} (x - 2) = 0

x^{3} = 0 or x - 2 = 0

x = 0 or x = 2

If x = 0

\frac{d y}{d x} = \frac{3 (0)^{2}}{y}

\frac{d y}{d x} = 0 which is not possible

So we take x = 2 and substitute it in y^{2} = 2 x^{3}

\begin{aligned} y^{2} = 2 (2)^{3} \\ y^{2} = 2 \times 8 \\ y^{2} = 16 \end{aligned}

y = 4 \dots . . {As y = x^{2}, - 4 as been discarded}

So,the required point is (2, 4)

Tangents and Normals Exercise 15 .1 Question 7 .

Answer: $The required points are (2, - 2) & (- 2, 2)$
Hint:

The slope of the tangent is the limit of \frac{Δ y}{Δ x} as Δ x approaches zero.

Given:

The curve x y + 4 = 0

Solution:

If a tangent line to the curve y = f (x) makes an angle θ with x -axis in the positive direction, then

\frac{d y}{d x} = The slope of\: tangent = \tan θ

x y + 4 = 0

Differentiating the above with respect to x

x \frac{d}{d x} (y) + y \frac{d}{d x} (x) + \frac{d}{d x} (4) = 0

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

∴ \frac{d}{d x} (constant) = 0

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

\frac{d y}{d x} = \frac{- y}{x} \to (1)

Also, \frac{d y}{d x} = \tan 45^{\circ} = 1 \to (2)

From (1)\&(2) we get

\begin{matrix} \frac{- y}{x} = 1 \\ x = - y \end{matrix}

Substituting x=-y in xy+4=0

\begin{aligned} x (- x) + 4 = 0 \\ - x^{2} + 4 = 0 \\ x^{2} = 4 \\ x = \pm 2 \end{aligned}

So when x = 2, y = - 2

and when x = - 2, y = 2

Thus the required points are (2, - 2) & (- 2, 2)

Tangents and Normals Exercise 15.1 Question 8 .

Answer: $The required point is (0, 0)$
Hint: $The slope of the tangent is the limit of \frac{Δ y}{Δ x} as Δ x approaches zero.$
Given: $The curve y = x^{2}$
Solution: $y = x^{2}$
Differentiating the above with respect to

x

∴ \frac{d}{d x} (x^{n}) = n x^{n - 1}

\frac{d y}{d x} = 2 x^{2 - 1}

\frac{d y}{d x} = 2 x \to (1)

Also given the slope of tangent is equal to the x-coordinate

\frac{d y}{d x} = x \to (2)

From (1)\&(2) we get

2 x = x

x = 0

Substituting this in y = x^{2}, we get

y = (0)^{2}

y = 0

So,the required point is (0, 0)

Tangents and Normals Exercise 15.1 Question 9 .

Answer: $The required point is (1, 0) & (1, 4)$
Hint: $The slope of the tangent is the limit of \frac{Δ y}{Δ x} as Δ x approaches zero.$
Given: $The curve is x^{2} + y^{2} - 2 x - 4 y + 1 = 0$
Solution: $x^{2} + y^{2} - 2 x - 4 y + 1 = 0$
$Differentiating the above with respect to x$
$∴ \frac{d}{d x} (x^{n}) = n x^{n - 1}, \frac{d}{d x} (constant) = 0$
$2 x^{2 - 1} + 2 y^{2 - 1} \frac{d y}{d x} - 2 - 4 \frac{d y}{d x} + 0 = 0$
$2 x + 2 y \frac{d y}{d x} - 2 - 4 \frac{d y}{d x} = 0$
$\frac{d y}{d x} (2 y - 4) = - 2 x + 2$
$\frac{d y}{d x} = \frac{- (x - 1)}{(y - 2)} \to (1)$
$\frac{d y}{d x} = The slope of tangent = \tan θ$
$Since the tangent is parallel to the x -axis$
$\frac{d y}{d x} = \tan (0) = 0 \to (2)$
From (1)&(2) we get
$\frac{- (x - 1)}{(y - 2)} = 0$
$- (x - 1) = 0$
$- x + 1 = 0$
$x = 1$
$Substituting x = 1 in x^{2} + y^{2} - 2 x - 4 y + 1 = 0, we get$
${(1)}^{2} + y^{2} - 2 (1) - 4 y + 1 = 0$
$1 + y^{2} - 2 - 4 y + 1 = 0$
$y^{2} - 4 y = 0$
$y (y - 4) = 0$
$y = 0 & y - 4 = 0$
$y = 0 & y = 4$
$So, the required point is (1, 0) & (1, 4)$

Tangents and Normals Exercise 15.1 Question 10 .

Answer: $The required point is (\frac{1}{2}, \frac{1}{4})$
Hint: $The slope of the tangent is the limit of \frac{Δ y}{Δ x} as Δ x approaches zero.$
Given: $The curve is y = x^{2}$
Solution: $y = x^{2}$
$Differentiating the above with respect to x$
$∴ \frac{d}{d x} (x^{n}) = n x^{n - 1}$
$\frac{d y}{d x} = 2 x^{2 - 1}$
$\frac{d y}{d x} = 2 x \to (1)$
$\frac{d y}{d x} = The slope of tangent = \tan θ$
$Since the tangent make an angle of 45^{\circ} with x -axis$
$\frac{d y}{d x} = \tan (45^{\circ}) = 1 \to (2)$
From (1)&(2) we get

2 x = 1

x = \frac{1}{2}

Substituting x = \frac{1}{2} in y = x^{2}, we get

y = {(\frac{1}{2})}^{2}

y = \frac{1}{4}

So, the required point is

(\frac{1}{2}, \frac{1}{4})

Tangents and Normals Exercise 15.1 Question 10

Answer: $The required points are (\frac{5}{3}, \frac{4}{3}) or (\frac{4}{3}, \frac{4}{3})$
Hint: $The slope of the tangent is the limit of \frac{Δ y}{Δ x} as Δ x approaches zero.$
Given: $The curve is y = 3 x^{2} - 9 x + 8$
Solution: $y = 3 x^{2} - 9 x + 8$
$Differentiating the above with respect to x$
$∴ \frac{d}{d x} (x^{n}) = n x^{n - 1}, \frac{d}{d x} (constant) = 0$
$\frac{d y}{d x} = 3 (2) x^{2 - 1} - 9$
$\frac{d y}{d x} = 6 x - 9 \to (1)$
$\frac{d y}{d x} = The slope of tangent = \tan θ$
$The tangent is equally inclined with the axis θ = \frac{π}{4} or \frac{- π}{4}$
$\frac{d y}{d x} = \tan (\frac{π}{4}) or \tan (- \frac{π}{4})$
$= 1 or - 1 \to (2)$
From(1)&(2) we get
$6 x - 9 = 1 or 6 x - 9 = - 1$
$6 x = 1 + 9 or 6 x = - 1 + 9$
$6 x = 10 or 6 x = 8$
$x = \frac{10}{6} or x = \frac{8}{6}$
$x = \frac{5}{3} or x = \frac{4}{3}$
Thus the requried points are

(\frac{5}{3}, \frac{4}{3}) or (\frac{4}{3}, \frac{4}{3})

Tangents and Normals Exercise 15.1 Question 12

Answer: $The required point is (1, 2)$
Hint: $The slope of the tangent is the limit of \frac{Δ y}{Δ x} as Δ x approaches zero.$
Given:
$y = 2 x^{2} - x + 1 \to (1)$
$y = 3 x + 4 \to (2)$
Solution: $Differentiating eqn(1) with respect to x$
$\begin{aligned} ∴ \frac{d}{d x} (x^{n}) = n x^{n - 1}, \frac{d}{d x} (constant) = 0 \\ \frac{d y}{d x} = 2 \times 2 (x)^{2 - 1} - 1 + 0 \\ \frac{d y}{d x} = 4 x - 1 \to (3) \end{aligned}$
$Again differentiating eqn (2) with respect to x$
$\frac{d y}{d x} = 3 \to (3)$
ATQ

The curve y = 2 x^{2} - x + 1 is the tangent parallel to the line y = 3 x + 4

\begin{aligned} 4 x - 1 = 3 \\ 4 x = 3 + 1 \\ 4 x = 4 \\ x = \frac{4}{4} = 1 \end{aligned}

Thus from(1)

y = 2

Hence the point is (1,2)

Tangents and Normals Exercise15.1 Question 13 .

Answer:

The required point is (1, 7)

Hint:

The slope of the tangent is the limit of \frac{Δ y}{Δ x} as Δ x approaches zero.

Given: $The curve is y = 3 x^{2} + 4$
Solution: $y = 3 x^{2} + 4$
$Differentiating the above with respect to x$
$∴ \frac{d}{d x} (x^{n}) = n x^{n - 1}, \frac{d}{d x} (constant) = 0$
$\frac{d y}{d x} = 3 \times (2) x^{2 - 1}$
$\frac{d y}{d x} = 6 x \to (1)$
$The slope m_{1} = \frac{d y}{d x} = 6 x$
$Now the given slope is m_{2} = \frac{- 1}{6}$
$We have tangent is perpendicular to tangent whose slope is \frac{- 1}{6}$
$m_{1} \times m_{2} = - 1$
$6 x \times \frac{- 1}{6} = - 1$
$x = 1$
From(1)

y = 7

Thue the required points is (1,7)

Tangents and Normals Exercise15.1 Question 14 .

Answer: The requried point are

(2, 3)

(- 2, - 3)

Hint:

The slope of the tangent is the limit of \frac{Δ y}{Δ x} as Δ x approaches zero.

GIven:
$Equation of the curve is x^{2} + y^{2} = 13 \to (1)$
$Equation of the line is 2 x + 3 y = 7 \to (2)$
Solution:
$Differentiating eqn(1) with respect to x$
$\begin{aligned} ∴ \frac{d}{d x} (x^{n}) = n x^{n - 1}, \frac{d}{d x} (constant) = 0 \\ 2 x^{2 - 1} + 2 y^{2 - 1} \frac{d y}{d x} = 0 \end{aligned}$
$\begin{aligned} 2 x + 2 y \frac{d y}{d x} = 0 \\ 2 y \frac{d y}{d x} = - 2 x \\ \frac{d y}{d x} = \frac{- x}{y} \end{aligned}$
$Slope m_{1} for (1) = \frac{d y}{d x} = \frac{- x}{y} \to (3)$
$Differentiating eqn(1) with respect to x$
$2 + 3 \frac{d y}{d x} = 0$
$3 \frac{d y}{d x} = - 2$
$\frac{d y}{d x} = \frac{- 2}{3}$
$Slope m_{2} = \frac{d y}{d x} = \frac{- 2}{3}$
ATQ
$m_{1} = m_{2} \dots . {as the tangents are parallel}$
$\frac{- x}{y} = \frac{- 2}{3}$
$x = \frac{2}{3} y$
$When put x = \frac{2}{3} y in eqn (1) x^{2} + y^{2} = 13$
${(\frac{2}{3} y)}^{2} + y^{2} = 13$
$\frac{4}{9} y^{2} + y^{2} = 13$
L.C.M.of 1ans 9 is 9
$\frac{4 y^{2} + 9 y^{2}}{9} = 13$
$\frac{13 y^{2}}{9} = 13$
$y^{2} = 9$
$y = \pm 3$
Put $y = \pm 3 in x = \frac{2}{3} y$
$x = \frac{2}{3} (\pm 3)$
$x = \pm 2$
Thus the requried points are

(2, 3)

(- 2, - 3)

Tangents and Normals Exercise 15.1 Question 15

Answer: The requried point are

(0, 0) & (2 a, - 2 a)

Hint:

The slope of the tangent is the limit of \frac{Δ y}{Δ x} as Δ x approaches zero.

Given:The Curve is

2 a^{2} y = x^{3} - 3 a x^{2} \to (1)

Solution:

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

Differentiating the above with respect to x

∴ \frac{d}{d x} (x^{n}) = n x^{n - 1}, \frac{d}{d x} (constant) = 0

2 a^{2} \frac{d y}{d x} = 3 x^{3 - 1} - 3 a (2) (x)^{2 - 1}

2 a^{2} \frac{d y}{d x} = 3 x^{2} - 6 a x

\frac{d y}{d x} = \frac{1}{2 a^{2}} [3 x^{2} - 6 a x]

Slope m_{1} = \frac{d y}{d x} = \frac{1}{2 a^{2}} [3 x^{2} - 6 a x] \to (2)

Also

Slope m_{2} = \frac{d y}{d x} = \tan θ

\tan 0^{\circ} = 0

[slope is parallel to x -axis]

m_{1} = m_{2}

\frac{1}{2 a^{2}} [3 x^{2} - 6 a x] = 0

3 x^{2} - 6 a x = 0

\begin{aligned} 3 x (x - 2 a) = 0 \\ 3 x = 0 or x - 2 a = 0 \\ x = 0 or x = 2 a \end{aligned}

When put x = 0 in eqn(1) 2 a^{2} y = x^{3} - 3 a x^{2}

\begin{aligned} 2 a^{2} y = (0)^{3} - 3 a (0)^{2} \\ 2 a^{2} y = 0 \\ y = 0 \end{aligned}

When put x = 2 a in eqn (1) 2 a^{2} y = x^{3} - 3 a x^{2}

\begin{aligned} 2 a^{2} y = (2 a)^{3} - 3 a (2 a)^{2} \\ 2 a^{2} y = 8 a^{3} - 3 a (4 a^{2}) \\ 2 a^{2} y = 8 a^{3} - 12 a^{3} \\ 2 a^{2} y = - 4 a^{3} \\ y = - 2 a \end{aligned}

Thus the requried points are

(0, 0) & (2 a, - 2 a)

Tangents and Normals Exercise 15.1 Question 16

Answer: The requried points are (3,2)
Hint:

The slope of the tangent is the limit of \frac{Δ y}{Δ x} as Δ x approaches zero.

Given:

Equation of the curve is y = x^{2} - 4 x + 5 \to (1)

Equation of the line is 2 y + x = 7 \to (2)

Solution:

Differentiating eqn(1) with respect to x

∴ \frac{d}{d x} (x^{n}) = n x^{n - 1}, \frac{d}{d x} (constant) = 0

\frac{d y}{d x} = 2 x^{2 - 1} - 4

\frac{d y}{d x} = 2 x - 4

Slope m_{1} = \frac{d y}{d x} = 2 x - 4 \to (3)

Differentiating eqn (2) with respect to x

2 \frac{d y}{d x} + 1 = 0

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

\frac{d y}{d x} = \frac{- 1}{2}

Slope m_{2} = \frac{d y}{d x} = \frac{- 1}{2}

We have given that slope(1) and slope (2) are perpendicular to each other

Hence,

\begin{aligned} m_{1} \times m_{2} = - 1 \\ (2 x - 4) \times (\frac{- 1}{2}) = - 1 \\ (2 x) (\frac{- 1}{2}) - (4) (\frac{- 1}{2}) = - 1 \end{aligned}

\begin{aligned} - x + 2 = - 1 \\ x = 2 + 1 \\ x = 3 \end{aligned}

When put x = 3 in eqn (1) y = x^{2} - 4 x + 5

\begin{aligned} y = (3)^{2} - 4 (3) + 5 \\ y = 9 - 12 + 5 \\ y = 2 \end{aligned}

Thus the required points are (3,2)

Tangents and Normals Exercise 15.1 Question 17 Sub Question 1 .

Answer: $The points at which the tangents are parallel to x -axis are (0, 5) & (0, - 5)$
Hint: $The slope of the tangent is the limit of \frac{Δ y}{Δ x} as Δ x approaches zero.$
Given: $\frac{x^{2}}{4} + \frac{y^{2}}{25} = 1 \to (1)$
Solution: $Differentiating eqn(1) with respect to x$
$\frac{d}{d x} (x^{n}) = n x^{n - 1}, \frac{d}{d x} (constant) = 0$
$\frac{2 x^{2 - 1}}{4} + \frac{2 y}{25} \frac{d y}{d x} = 0$
$\frac{2 x}{4} + \frac{2 y}{25} \frac{d y}{d x} = 0$
$\frac{x}{2} + \frac{2 y}{25} \frac{d y}{d x} = 0$
$\frac{2 y}{25} \frac{d y}{d x} = \frac{- x}{2}$
$\frac{d y}{d x} = \frac{- x}{2} \times \frac{25}{2 y}$
$\frac{d y}{d x} = \frac{- 25 x}{4 y}$
Now the tangent is parallel to the x-axis if the slope of the tangent is zero

\frac{- 25 x}{4 y} = 0

This is possible if x = 0

Then, \frac{x^{2}}{4} + \frac{y^{2}}{25} = 1 for x = 0

\begin{aligned} \frac{0}{4} + \frac{y^{2}}{25} = 1 \\ y^{2} = 25 \\ y = \pm 5 \end{aligned}

Substitute y = \pm 5 in \frac{- 25 x}{4 y}

\begin{array}{ll} Put y = 5 & , Put y = - 5 \\ \frac{- 25 x}{4 (5)} = 0 & , \frac{- 25 x}{4 (- 5)} = 0 \\ \frac{- 5 x}{4} = 0 & , \frac{5 x}{4} = 0 \end{array}

x = 0, x = 0

Thus the points at which the tangents are parallel to x-axis are 0,5&0,-5

Tangents and Normals Exercise 15.1 Question 17 Sub Question 2 .

Answer: $The points at which the tangents are parallel to y -axis are (2, 0) & (- 2, 0)$
Hint: $The slope of the tangent is the limit of \frac{Δ y}{Δ x} as Δ x approaches zero.$
Given: $\frac{x^{2}}{4} + \frac{y^{2}}{25} = 1 \to (1)$
Solution: $Differentiating eqn(1) with respect to x$
$\frac{d}{d x} (x^{n}) = n x^{n - 1}, \frac{d}{d x} (constant) = 0$
$\frac{2 x^{2 - 1}}{4} + \frac{2 y}{25} \frac{d y}{d x} = 0$
$\frac{2 x}{4} + \frac{2 y}{25} \frac{d y}{d x} = 0$
$\begin{aligned} \frac{x}{2} + \frac{2 y}{25} \frac{d y}{d x} = 0 \\ \frac{2 y}{25} \frac{d y}{d x} = \frac{- x}{2} \\ \frac{d y}{d x} = \frac{- x}{2} \times \frac{25}{2 y} \\ \frac{d y}{d x} = \frac{- 25 x}{4 y} & \frac{d x}{d y} = \frac{- 4 y}{25 x} \end{aligned}$
Now the tangent is parallel to the y-axis if the slope of the normal is zero
Slope of normal=

\frac{- 1}{Slope of tangent} \Rightarrow \frac{- 1}{\frac{- 25 x}{4 y}} = 0

\frac{4 y}{25 x} = 0

This is possible if y = 0

Then, \frac{x^{2}}{4} + \frac{y^{2}}{25} = 1 for y = 0

\frac{x^{2}}{4} + \frac{0}{25} = 1

x^{2} = 4

x = \pm 2

Thus $The points at which the tangents are parallel to y -axis are (2, 0) & (- 2, 0)$

Tangents and Normals Exercise 15.1 Question 18 Sub Question 1 .

Answer:

The points at which the tangents are parallel to x -axis are (1, 2) & (1, - 2)

Hint:

The slope of the tangent is the limit of \frac{Δ y}{Δ x} as Δ x approaches zero.

Given:

x^{2} + y^{2} - 2 x - 3 = 0 \to (1)

Solution:

Differentiating eqn (1) with respect to x

\frac{d}{d x} (x^{n}) = n x^{n - 1}, \frac{d}{d x} (constant) = 0

2 x^{2 - 1} + 2 y^{2 - 1} \frac{d y}{d x} - 2 = 0

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

\begin{aligned} 2 y \frac{d y}{d x} = 2 - 2 x \\ 2 y \frac{d y}{d x} = 2 (1 - x) \\ \frac{d y}{d x} = \frac{1 - x}{y} \end{aligned}

Now the tangent is parallel to the x-axis if the slope of the tangent is zero

\begin{aligned} \frac{1 - x}{y} = 0 \\ 1 - x = 0 \\ x = 1 \end{aligned}

substitute x = 1 in x^{2} + y^{2} - 2 x - 3 = 0

(1)^{2} + y^{2} - 2 (1) - 3 = 0

1 + y^{2} - 2 - 3 = 0

1 + y^{2} - 5 = 0

y^{2} = 5 - 1

y^{2} = 4

y = \pm 2

Thus the points at which the tangents are parallel to x -axis are (1, 2) & (1, - 2)

Tangents and Normals Exercise 15.1 Question 18 Sub Question 2 .

Answer: $The points at which the tangents are parallel to y -axis are (- 1, 0) & (3, 0)$
Hint: $The slope of the tangent is the limit of \frac{Δ y}{Δ x} as Δ x approaches zero.$
Given: $x^{2} + y^{2} - 2 x - 3 = 0 \to (1)$
Solution: $Differentiating eqn(1) with respect to x$
$\begin{aligned} \frac{d}{d x} (x^{n}) = n x^{n - 1}, \frac{d}{d x} (constant) = 0 \\ 2 x^{2 - 1} + 2 y^{2 - 1} \frac{d y}{d x} - 2 = 0 \\ 2 x + 2 y \frac{d y}{d x} = 2 \\ 2 y \frac{d y}{d x} = 2 - 2 x \\ 2 y \frac{d y}{d x} = 2 (1 - x) \\ \frac{d y}{d x} = \frac{1 - x}{y} \end{aligned}$
$\begin{aligned} Now the tangent is parallel to the y -axis if the slope of the normal is zero \\ Slope of normal = \frac{- 1}{Slope of tangent} \Rightarrow \frac{- 1}{\frac{1 - x}{y}} = \frac{- y}{1 - x} = 0 \end{aligned}$
But
$\begin{aligned} x^{2} + y^{2} - 2 x - 3 = 0 for y = 0 \\ x^{2} + 0 - 2 x - 3 = 0 \\ x^{2} - 2 x - 3 = 0 \\ x^{2} - 3 x + x - 3 = 0 \\ x (x - 3) + 1 (x - 3) = 0 \\ x + 1 = 0 & x - 3 = 0 \\ x = - 1 & x = 3 \end{aligned}$
Thus $The points at which the tangents are parallel to y -axis are (- 1, 0) & (3, 0)$

Tangents and Normals Exercise 15.1 Question 19 Sub Question 1 .

Answer:

The points at which the tangents are parallel to x -axis are (0, 4) & (0, - 4)

Hint:

The slope of the tangent is the limit of \frac{Δ y}{Δ x} as Δ x approaches zero.

Given:

\frac{x^{2}}{9} + \frac{y^{2}}{16} = 1 \to (1)

Solution:

Differentiating eqn(1) with respect to x

\begin{aligned} \frac{d}{d x} (x^{n}) = n x^{n - 1}, \frac{d}{d x} (constant) = 0 \\ \frac{2 x}{9} + \frac{2 y}{16} \frac{d y}{d x} = 0 \\ \frac{d y}{d x} = \frac{- 16}{9} \frac{x}{y} \end{aligned}

Now the tangent is parallel to the x -axis if the slope of the tangent is zero

\frac{- 16}{9} \frac{x}{y} = 0 which is possible if x = 0

Then,

\frac{x^{2}}{9} + \frac{y^{2}}{16} = 1 for x = 0

\begin{aligned} y^{2} = 16 \\ y = \pm 4 \end{aligned}

Thus the points at which the tangents are parallel to x -axis are (0, 4) & (0, - 4)

Tangents and Normals Exercise 15.1 Question 19 Sub Question 2 .

Answer: $The points at which the tangents are parallel to y -axis are (3, 0) & (- 3, 0)$

Hint: $The slope of the tangent is the limit of \frac{Δ y}{Δ x} as Δ x approaches zero.$
Given: $\frac{x^{2}}{9} + \frac{y^{2}}{16} = 1 \to (1)$
Solution: $Differentiating eqn(1) with respect to x$
$\begin{aligned} \frac{d}{d x} (x^{n}) = n x^{n - 1}, \frac{d}{d x} (constant) = 0 \\ \frac{2 x}{9} + \frac{2 y}{16} \frac{d y}{d x} = 0 \\ \frac{d y}{d x} = \frac{- 16}{9} \frac{x}{y} \end{aligned}$
$Now the tangent is parallel to the y -axis if the slope of the normal is zero$
$slope of the normal = \frac{- 1}{(\frac{- 16 x}{9 y})} \Rightarrow \frac{9 y}{16 x} = 0 which is possible if y = 0$
Then,
$\frac{x^{2}}{9} + \frac{y^{2}}{16} = 1 for y = 0$
$x^{2} = 9$
$x = \pm 3$
$Thus the points at which the tangents are parallel to y -axis are (3, 0) & (- 3, 0)$

Tangents and Normals Exercise 15.1 Question 20 .

Answer:
Hence Proved that the tangents to the curve

y = 7 x^{3} + 11 at the points x = 2 and x = - 2 are parallel

Hint:

The slope of the tangent is the limit of \frac{Δ y}{Δ x} as Δ x approaches zero.

Given:

The equation of the curve is y = 7 x^{3} + 11 \to (1)

Solution:

Differentiating eqn(1) with respect to x

\begin{aligned} \frac{d}{d x} (x^{n}) = n x^{n - 1}, \frac{d}{d x} (constant) = 0 \\ \frac{d y}{d x} = 7 (3) x^{3 - 1} + 0 \\ \frac{d y}{d x} = 21 x^{2} \end{aligned}

The slope of the tangent to a curve at (x_{0}, y_{0}) is \frac{d y}{d x (x_{0}, y_{0})}

Therefore the slope of the tangent at the point where x = 2 is given by

\frac{d y}{d x_{x = 2}} = 21 (2)^{2}

\begin{aligned} = 21 \times 4 \\ = 84 \end{aligned}

and the slope of the tangent at the point where x = - 2 is given by

\frac{d y}{d x_{x} = - 2} = 21 (- 2)^{2}

\begin{aligned} = 21 \times 4 \\ = 84 \end{aligned}

It is observed that the slopes of the tangent at the points where x=2 and x=-2 are equal.
Hence, the two tangents are parallel

Tangents and Normals Exercise 15.1 Question 21 .

Answer: The requried point is $(0, 0) or (\frac{1}{3}, \frac{1}{27})$
Hint: $The slope of the tangent is the limit of \frac{Δ y}{Δ x} as Δ x approaches zero.$
Given: $The equation of the curve is y = x^{3} \to (1)$
Solution: $Differentiating eqn(1) with respect to x$
$\begin{aligned} \frac{d}{d x} (x^{n}) = n x^{n - 1}, \frac{d}{d x} (constant) = 0 \\ \frac{d y}{d x} = 3 x^{2} \to (2) \end{aligned}$
$Slope of the tangent to (1) is$
$m_{1} = \frac{d y}{d x} = 3 x^{2}$
$Also, given that slope of the tangent is parallel to x -coordinate of the point$
$m_{2} = \frac{d y}{d x} = x \to (3)$
From (2) & (3)
$m_{1} = m_{2}$
$3 x^{2} = x$
$3 x^{2} - x = 0$
$x (3 x - 1) = 0$
$x = 0 or 3 x - 1 = 0$
$x = 0 or x = \frac{1}{3}$
$Substitute x = 0 in y = x^{3}$
$y = 0$
$Substitute x = \frac{1}{3} in y = x^{3}$
$y = {(\frac{1}{3})}^{3}$
$y = \frac{1}{27}$
$y = 0 or \frac{1}{27}$
$Thus the required point is (0, 0) or (\frac{1}{3}, \frac{1}{27})$

Tangents and Normals Exercise 15.2 Question 3 Sub Question 2 .

Answer: Equation of tangent $2 x - y + 1 = 0$
Equation of normal

x + 2 y - 7 = 0

HINTS:
Putting

x = 1

in the given equation and differentiate it.
GIVEN:
$y = x^{4} - 6 x^{3} + 13 x^{2} - 10 x + 5 at x = 1$
Solution:
$y = 1 - 6 + 13 - 10 + 5 = 3$
So, $(x_{1}, y_{1}) = (1, 3)$
Now,
$\begin{aligned} y = x^{4} - 6 x^{3} + 13 x^{2} - 10 x + 5 \\ \frac{d y}{d x} = 4 x^{3} - 18 x^{2} + 26 x - 10 \end{aligned}$
Slope of tangent ,

\begin{aligned} m & = {(\frac{d y}{d x})}_{(1, 3)} \\ = 4 - 18 + 26 - 10 \\ = 2 \end{aligned}

Equation of tangent is,

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

Equation of Normal is,

$\begin{aligned} y - y_{1} = - \frac{1}{m} (x - x_{1}) \\ \Rightarrow y - 3 = - \frac{1}{2} (x - 1) \\ \Rightarrow 2 y - 6 = - x + 1 \\ \Rightarrow x + 2 y - 7 = 0 \end{aligned}$