Question

The area of the rectangle formed by the perpendiculars from the centre of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}$= 1 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. $\frac{a^{2} + b^{2}}{ab(a^{2}–b^{2})}$

Answer 1

Answer: (A)

$\frac{(a^{2}–b^{2})ab}{a^{2} + b^{2}}$

Answer 2

Equation of the tangent at $\frac{\pi}{4}$ is

$\frac{x\left( \frac{1}{\sqrt{2}} \right)}{a} + \frac{y\left( \frac{1}{\sqrt{2}} \right)}{b}$= 1

i.e., $\frac{x}{a} + \frac{y}{b}$– Ö2 = 0 ... (1)

Equation of the normal at $\frac{\pi}{4}$ is

$\frac{x}{b}–\frac{y}{a} = \frac{a}{b\sqrt{2}}–\frac{b}{a\sqrt{2}}$... (2)

p1 = length of the perpendicular from the centre to the tangent

= $\left| \frac{–\sqrt{2}}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}}}} \right|$ = $\frac{\sqrt{2}ab}{\sqrt{a^{2} + b^{2}}}$

p2 = length of the perpendicular from the centre to the normal

= $\left| \frac{\frac{a}{b\sqrt{2}}–\frac{b}{a\sqrt{2}}}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}}}} \right|$ = $\left| \frac{a^{2}–b^{2}}{\sqrt{2}\sqrt{a^{2} + b^{2}}} \right|$

Area of the rectangle = p1p2 = $\frac{ab(a^{2}–b^{2})}{a^{2} + b^{2}}$