JEE Main & Advanced
Sample Paper
JEE Main Sample Paper-15
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