Question

Let $P$ be a variable point on the ellipse $\frac{x^2}{100}+\frac{y^2}{64}=1$ with foci $F_1$ and $F_2$. If $A$ is the area of triangle $P{F}_1{F}_2$, then the maximum possible value of $A$ is

• A. $120$
• B. $48$
• C. $152$
• D. $24$

Answer 1

The correct option is B $48$

$a=10,b=8$
$\therefore c=\sqrt{{10}^2-8^2}=6$
Any point $P$ on the ellipse is $P(10\cos \theta ,8\sin \theta )$.

$A=\frac{1}{2}(\text{Base}\times \text{height})\Rightarrow A=\frac{1}{2}(F_1{F}_2\times PM)\Rightarrow A=\frac{1}{2}(2\times 6\times 8\sin \theta )\Rightarrow A=48\sin \theta$
For maximum value of $A,\theta =\frac{\pi }{2}$
and $A_{max}=48$