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 $\frac{\pi }{4}$ is

• A. $\frac{(a^2-b^2)ab}{a^2+b^2}$
• B. $\frac{(a^2+b^2)ab}{a^2-b^2}$
• C. $\frac{(a^2-b^2)}{ab(a^2+b^2)}$
• D. $\frac{a^2+b^2}{(a^2-b^2)ab}$

Answer 1

The correct option is A $\frac{(a^2-b^2)ab}{a^2+b^2}$

Let equation of ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Equation of tangent at P $\left(acos\frac{\pi }{4},bsin\frac{\pi }{4}\right)$ is $\frac{x}{a}+\frac{y}{b}=\sqrt{2}$

Equation of normal at P is
$\sqrt{2}ax-\sqrt{2}by=a^2-b^2$

Now
$OT=\left|\frac{-\sqrt{2}ab}{\sqrt{a^2+b^2}}\right|$
And $ON=\left|\frac{-(a^2-b^2)}{\sqrt{2}\sqrt{a^2+b^2}}\right|$

Hence area is $\frac{(a^2-b^2)ab}{a^2+b^2}$