EllipseHardBloom L3
Question
The area of the rectangle formed by the perpendiculars from the centre of the standard ellipse to the tangent and normal at its point whose eccentric angle is $\pi/4$, is :
Options
A.$\frac{\left( a^{2} - b^{2} \right)ab}{a^{2} + b^{2}}$
B.$\frac{\left( a^{2} - b^{2} \right)}{\left( a^{2} + b^{2} \right)ab}$
C.$\frac{\left( a^{2} - b^{2} \right)}{ab\left( a^{2} + b^{2} \right)}$
D.$\frac{a^{2} + b^{2}}{\left( a^{2} - b^{2} \right)ab}$
Solution
Sol.(a) Equation of normal at $P$,
$$\begin{matrix} ax\sqrt{2} - by\sqrt{2} = a^{2} - b^{2} \\ P_{1} = \frac{a^{2} - b^{2}}{\sqrt{2\left( a^{2} + b^{2} \right)}} \end{matrix}$$
Equation of tangent at $P$ is
$$\begin{matrix} & \frac{x}{\sqrt{2}a} + \frac{y}{\sqrt{2}b} = 1 \\ P_{2} = & \frac{1}{\sqrt{\frac{1}{2a^{2}} + \frac{1}{2b^{2}}}} = \frac{\sqrt{2}ab}{\sqrt{a^{2} + b^{2}}} \\ P_{1}P_{2} = & \frac{ab\left( a^{2} - b^{2} \right)}{\left( a^{2} + b^{2} \right)} \end{matrix}$$
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