question_answer

The area of the rectangle formed by the perpendicular from centre of \[\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{4}=1,\]to the tangent and normal at the point whose eccentric angle is \[\frac{\pi }{4}\]equals

A) \[\frac{30}{13}\]                                              

B) \[\frac{30}{11}\]

C) \[\frac{13}{30}\]                                              

D) \[\frac{26}{11}\]

Correct Answer: A

Solution :

Area of rectangle \[OQPN=(ON)\,(OQ)\] \[=\left( \frac{6\sqrt{2}}{\sqrt{13}} \right)\left( \frac{5}{\sqrt{26}} \right)=\frac{30}{13}\]sq. units