The line 2x + y = 3 cuts the ellipse 4 x ^ 2 + y ^ 2 = 5 at P and Q . If θ is the acute angle between normal at P and tangent at Q , then θ is equal to

Correct answer: C. ^ - 1 ( 3 34 ) Solving y = 3 - 2x and 4 x ^ 2 + y ^ 2 = 5 , we get the intersection point P((1)/(2), 2) and Q(1, 1) The equation of tangent at these points are 4x((1)/(2)) + 2y = 5 and 4(x)(1) + y(1) = 5 ⇒ Slope of tangents at P and Q are - 1 and - 4 respectively ⇒ Slope of normal at P is 1 ⇒ tan θ = |(1 - (- 4))/(1 + 1(- 4))| = (5)/(3) ⇒ cos θ = 3 34

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Ellipse

NTA Abhyas 2020 · Mathematics · Ellipse

The line 2x + y = 3 cuts the ellipse 4 x ^ 2 + y ^ 2 = 5 at P and Q...

The line

2x + y = 3

cuts the ellipse

4{x}^{2} + {y}^{2} = 5

P

and

Q

. If

\theta

is the acute angle between normal at

P

and tangent at

Q

, then

\theta

is equal to

Options

A $\tan^{- 1} \left(\dfrac{4}{5}\right)$
B $\sin^{- 1} \left(\dfrac{3}{\sqrt{34}}\right)$
C $\cos^{- 1} \left(\dfrac{3}{\sqrt{34}}\right)$
D $\cot^{- 1} \left(\dfrac{3}{4}\right)$

Correct answer

C. $\cos^{- 1} \left(\dfrac{3}{\sqrt{34}}\right)$

Step-by-step solution

Solving

y = 3 - 2x

and

4{x}^{2} + {y}^{2} = 5

, we get the intersection point $

P\left(\dfrac{1}{2}, 2\right)

and

Q\left(1, 1\right)

The equation of tangent at these points are

4x\left(\dfrac{1}{2}\right) + 2y = 5

and

4\left(x\right)\left(1\right) + y\left(1\right) = 5

⇒

Slope of tangents at

P

and

Q

are

- 1

and

- 4

respectively

⇒

Slope of normal at

P

1

$⇒ tan \theta = \left|\dfrac{1 - \left(- 4\right)}{1 + 1\left(- 4\right)}\right| = \dfrac{5}{3} ⇒ cos \theta = \dfrac{3}{\sqrt{34}}

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