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  • If the normal at the point  to the ellipse  <img alt="\frac{\math

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

Option: 1

2 / 3


Option: 2

3 / 2


Option: 3

-2 / 3


Option: 4

-3 / 2


Answers (1)

best_answer

Equation of normal at (\mathrm a \cos \theta, \mathrm b \sin \theta)  is
\mathrm {ax} \sec \theta- \mathrm {by} \operatorname{cosec} \theta=\mathrm a^2-\mathrm b^2
Now this normal passes through ( \mathrm{a} \cos 2 \theta, \mathrm{b} \sin 2 \theta ), so

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

Hence (2) is the correct answer.

Posted by

Suraj Bhandari

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