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

120

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48

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152

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24

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Solution

The correct option is B $48$

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

$A = \frac{1}{2} (Base \times height) \Rightarrow A = \frac{1}{2} (F_{1} F_{2} \times P M) \Rightarrow A = \frac{1}{2} (2 \times 6 \times 8 sin θ) \Rightarrow A = 48 sin θ$
For maximum value of $A, θ = \frac{π}{2}$
and $A_{max} = 48$

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