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Question: Answered & Verified by Expert
Consider the conic $\mathrm{ex}^{2}+\pi y^{2}-2 \mathrm{e}^{2} \mathrm{x}-2 \pi^{2} \mathrm{y}+\mathrm{e}^{3}+\pi^{3}=\pi \mathrm{e}$. Suppose $\mathrm{P}$ is any point on the conic and $\mathrm{S}_{1}, \mathrm{~S}_{2}$ are the foci of the conic, then the maximum value of $\left(\mathrm{PS}_{1}+\mathrm{PS}_{2}\right)$ is $-$
MathematicsEllipseJEE Main
Options:
  • A $\pi \mathrm{e}$
  • B $\sqrt{\pi \mathrm{e}}$
  • C $2 \sqrt{\pi}$
  • D $2 \sqrt{\mathrm{e}}$
Solution:
2852 Upvotes Verified Answer
The correct answer is: $2 \sqrt{\pi}$
$e x^{2}+\pi y^{2}-2 e^{2} x-2 \pi^{2} y+e^{3}+\pi^{3}=\pi e$ $e\left(x^{2}-2 e x+e^{2}\right)+\pi\left(y^{2}-2 \pi y+\pi^{2}\right)=\pi e$ $\frac{(x-e)^{2}}{\pi}+\frac{(y-\pi)^{2}}{e}=1$
$a^{2}=\pi \Rightarrow a=\Gamma \pi \quad \quad \pi>e$
$P S_{1}+P S_{2}=2 a$ Major axis is $\|$ to axis
$P S_{1}+P S_{2}=2 \cdot F \bar{\pi}$

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