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Question

$The area of the rectangle formed by the perpendiculars from centre of \frac{x^{2}}{9} + \frac{y^{2}}{4} = 1 to the tangent and normal at$
$the point whose eccentric angle is \frac{π}{4} is A . The value of 13A is$

Moderate

Solution

$Equation of tangent at point P (\frac{π}{4}) is$
$\begin{array}{l} \frac{x}{3} \cos \frac{π}{4} + \frac{y}{2} \sin \frac{π}{4} = 1 \\ o r 2 x + 3 y - 6 \sqrt{2} = 0 \end{array}$
Equation of normal at P is
$3 \sqrt{2} x - 2 \sqrt{2} y - 5 = 0$
$Distance of centre from the tangent is \frac{6 \sqrt{2}}{\sqrt{13}} .$
$Distance of centre from the normal is \frac{5}{\sqrt{2} \sqrt{13}} .$
$∴ Area of required rectangle = \frac{6 \sqrt{2}}{\sqrt{13}} \times \frac{5}{\sqrt{2} \sqrt{13}} = \frac{30}{13} sq. units.$

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