A normal is drawn to the ellipse (x2)/(9) + y^ 2 = 1 at the point (3cos θ, sin θ) where 0 < θ < (π)/(2) . If N is the foot of the perpendicular from the origin~ O ~to the normal such that ON = 2 , then θ is equal to

Correct answer: D. (π)/(6) Equation of the normal at& (3cosθ, sinθ) is (3x)/(cos θ) - (y)/(sin θ) = 8 So,~ ON = | - 8 9 ^ 2 θ + cosec ^ 2 θ | = 2 ~(given) ⇒ 16 = 9(1 + ^ 2 θ) + (1 + ^ 2 θ) ⇒ 6 = 6 + (3tan θ - cot θ)2 ⇒ 3tan θ - cot θ = 0 Hence,~ ^ 2 θ = (1)/(3) ⇒ θ = (π)/(6)

NTA Abhyas 2020

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

NTA Abhyas 2020 · Mathematics · Ellipse

A normal is drawn to the ellipse (x2)/(9) + y^ 2 = 1 at the point...

A normal is drawn to the ellipse

\dfrac{x2}{9} + y^{2} = 1

at the point

\left(3cos \theta, sin \theta\right)

where

0 < \theta < \dfrac{\pi}{2} .

N

is the foot of the perpendicular from the origin~

O

~to the normal such that

ON = 2

, then

\theta

is equal to

Options

A $\dfrac{\pi}{4}$
B $\dfrac{\pi}{12}$
C $\dfrac{\pi}{3}$
D $\dfrac{\pi}{6}$

Correct answer

D. $\dfrac{\pi}{6}$

Step-by-step solution

Equation of the normal at& $\left(3cos\theta, sin\theta\right)$ is $\dfrac{3x}{cos \theta} - \dfrac{y}{sin \theta} = 8$
So,~ $ON = \left|\dfrac{- 8}{9\sec^{2} \theta + {cosec}^{2} \theta}\right| = 2$ ~(given)
$⇒ 16 = 9\left(1 + \tan^{2} \theta\right) + \left(1 + \cot^{2} \theta\right) ⇒ 6 = 6 + \left(3tan \theta - cot \theta\right)2$

$⇒ 3tan \theta - cot \theta = 0$
Hence,~ $\tan^{2} \theta = \dfrac{1}{3} ⇒ \theta = \dfrac{\pi}{6}$

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