Question

Let P be a variable point on the ellipse $\frac{x^{2}}{a^{2}}$+ $\frac{y^{2}}{b^{2}}$= 1 with foci S₁ and S₂. If A be the area

ofDPS₁S₂, then the maximum value of A:

• A. ab sin q
• B. abe
• C. a sin q
• D. b sin q

Answer 1

Answer: (B)

abe

Answer 2

Let P (a cos q, b sin q) be the variable point on the ellipse$\frac{x^{2}}{a^{2}}$+ $\frac{y^{2}}{b^{2}}$= 1. Then,

A = Area of DPS₁S₂,

= $\frac{1}{2}$ $\left| \begin{matrix} a\cos\theta & b\sin\theta & 1 \ ae & 0 & 1 \

• ae & 0 & 1 \end{matrix} \right|$=$\frac{1}{2}$b sin q × 2 ae

= abe sin q

= Area = abe sin q, which is maximum

when q = $\frac{\pi}{2}$

\ A_max = abe