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  • The tangent and normal to the ellipse at a point <img alt="\math

The tangent and normal to the ellipse \mathrm{x^2+4 y^2=4} at a point \mathrm{P(\theta)} in second quadrant, meet the major axis in Q and R respectively. If \mathrm{Q R=2}, then \cos \theta is equal to
 

Option: 1

\frac{-2 }{3}
 


Option: 2

\frac{2}{3}


Option: 3

\frac{1 }{3}


Option: 4

none of these


Answers (1)

best_answer

Ellipse \mathrm{\equiv \frac{x^2}{4}+y^2=1}

Let \mathrm{ P(\theta) \equiv(2 \cos \theta, \sin \theta)}

Equation of the tangent at

\mathrm{P: \frac{x \cos \theta}{2}+y \sin \theta=1 }
\begin{aligned} &\mathrm{ =Q \equiv(2 \sec \theta, 0) }\\ &\mathrm{ \text { Equation of the normal at } P: 2 x \sec \theta-\mathrm{y} \operatorname{cosec} \theta=3 \Rightarrow R=\left(\frac{3}{2} \cos \theta, 0\right) }\\ &\mathrm{ \text { Therefore } Q R=\left|2 \sec \theta-\frac{3}{2} \cos \theta\right|=2 \Rightarrow\left|\frac{4-3 \cos ^2 \theta}{2 \cos \theta}\right|=2 }\\ &\mathrm{ \qquad 16+9 \cos ^4 \theta-24 \cos ^2 \theta=16 \cos ^2 \theta \text { or } 9 \cos ^2 \theta-40 \cos ^2 \theta+16=0 }\\ &\mathrm{ \text { or } 9 \cos ^4 \theta-36 \cos ^2 \theta-4 \cos ^2 \theta+16=0 \text { or }\left(9 \cos ^2 \theta-4\right)\left(\cos ^2 \theta-4\right)=0 }\\ &\mathrm{ =\cos ^2 \theta=\frac{4}{9}}\\ \end{aligned}
\mathrm{=\cos \theta= \pm \frac{2}{3} \Rightarrow \cos \theta=-\frac{2}{3} }  as the point is in quadrant second.

Hence (a), is the correct answer.

Posted by

Ritika Harsh

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