Question

The area of the rectangle formed by the perpendiculars from the centre of the ellipse to the tangent and normal at the point-whose eccentric angle is $\pi/4$, is

• A. $\left( \frac{a^{2} - b^{2}}{a^{2} + b^{2}} \right)ab$
• B. $\left( \frac{a^{2} + b^{2}}{a^{2} - b^{2}} \right)ab$
• C. $\frac{1}{ab}\left( \frac{a^{2} - b^{2}}{a^{2} + b^{2}} \right)ab$
• D. $\frac{1}{ab}\left( \frac{a^{2} + b^{2}}{a^{2} - b^{2}} \right)ab$

Answer 1

Answer: (A)

$\left( \frac{a^{2} - b^{2}}{a^{2} + b^{2}} \right)ab$

Answer 2

The given point is ($a\cos\pi/4,b\sin\pi/4)$ i.e. $\left( \frac{a}{\sqrt{2}},\frac{b}{\sqrt{2}} \right)$.

So, the equation of the tangent at this point is $\frac{x}{a} + \frac{y}{b} = \sqrt{2}$ ......(i)

∴ $p_{1} =$ length of the perpendicular form (0, 0) on (i) = $\left| \frac{\frac{0}{a} + \frac{0}{b} - \sqrt{2}}{\sqrt{1/a^{2} + 1/b^{2}}} \right| = \frac{\sqrt{2}ab}{\sqrt{a^{2} + b^{2}}}$Equation of the normal at $\left( \frac{a}{\sqrt{2}},\frac{b}{\sqrt{2}} \right)$is $\frac{a^{2}x}{a/\sqrt{2}} - \frac{b^{2}y}{b/\sqrt{2}} = a^{2} - b^{2}$⇒

$\sqrt{2}ax - \sqrt{2}by = a^{2} - b^{2}$ .....(ii)

Therefore, $p_{2} =$length of the perpendicular form (0, 0) on (ii) $= \frac{a^{2}b^{2}}{\sqrt{(\sqrt{2a})^{2} + ( - \sqrt{2b})^{2}}} = \frac{a^{2} - b^{2}}{\sqrt{2(a^{2} + b^{2})}}$

So, area of the rectangle

$= p_{1}p_{2} = \frac{\sqrt{2}ab}{\sqrt{a^{2} + b^{2}}} \times \frac{a^{2} - b^{2}}{\sqrt{2(a^{2} + b^{2})}} = \left( \frac{a^{2} - b^{2}}{a^{2} + b^{2}} \right)ab$