Question

The area of the rectangle (in sq. units) formed by the perpendiculars drawn from the centre of the ellipse having major axis and minor axis lengths as $2a$ and $2b$ units respectively to the tangent and normal at a 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 be
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b)$
$P\left(\theta \right)=P\left(\frac{\pi }{4}\right)$
$\Rightarrow P=(\frac{a}{\sqrt{2}},\frac{b}{\sqrt{2}})$
Tangent at $P:\frac{x}{a\sqrt{2}}+\frac{y}{b\sqrt{2}}=1$
Normal at $P:\sqrt{2}ax-\sqrt{2}by=a^2-b^2$
Distance of Normal from $O(0,0)$
$d_1=\frac{a^2-b^2}{\sqrt{2a^2+2b^2}}$ units
Distance of tangent from $O(0,0)$
$d_2=\frac{1}{\sqrt{\frac{1}{2a^2}+\frac{1}{2b^2}}}$
$d_2=\frac{\sqrt{2}ab}{\sqrt{a^2+b^2}}$ units
Area of rectangle $=d_1\cdot d_2$
$=\frac{a^2-b^2}{\sqrt{2a^2+2b^2}}\cdot \frac{\sqrt{2}ab}{\sqrt{a^2+b^2}}$
$=\frac{(a^2-b^2)ab}{a^2+b^2}$ sq.units