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If the normal at the point  \mathrm{P(\theta)} to the ellipse  \mathrm{\frac{x^2}{14}+\frac{y^2}{5}=1}  intersects it again at 

the point \mathrm{Q(2 \theta)} , then \mathrm{\cos \theta} is equal to 

Option: 1

2/3


Option: 2

-2/3


Option: 3

3/2


Option: 4

-3/2


Answers (1)

best_answer

Equation of normal at \mathrm{(a \cos \theta, b \sin \theta)} is

 

\mathrm{a x \sec \theta-b y \operatorname{cosec} \theta=a^2-b^2}

 

Now this normal passes through \mathrm{(a \cos 2 \theta, b \sin 2 \theta)} , so

 

\mathrm{a \cdot a^{\frac{\cos 2 \theta}{\cos \theta}}-b \cdot b^{\frac{\sin 2 \theta}{\sin \theta}}=a^2-b^2}

 

\mathrm{\begin{aligned} & \Rightarrow 18 \cos ^2 \theta-9 \cos \theta-14=0 \\\\ & \Rightarrow \cos \theta=-\frac{2}{3} \text { or } \cos \theta=\frac{7}{6} \text { (Not possible) } \end{aligned}}

Posted by

Divya Prakash Singh

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