JEE MAIN - Mathematics (2025 - 4th April Evening Shift - No. 10)
The centre of a circle C is at the centre of the ellipse $\mathrm{E}: \frac{x^2}{\mathrm{a}^2}+\frac{y^2}{\mathrm{~b}^2}=1, \mathrm{a}>\mathrm{b}$. Let C pass through the foci $F_1$ and $F_2$ of E such that the circle $C$ and the ellipse $E$ intersect at four points. Let P be one of these four points. If the area of the triangle $\mathrm{PF}_1 \mathrm{~F}_2$ is 30 and the length of the major axis of $E$ is 17 , then the distance between the foci of $E$ is :
12
26
13
$\frac{13}{2}$
Explanation
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$$\begin{aligned} & x^2+\frac{a^2 y^2}{b^2}=a^2 \\ & \Rightarrow y^2\left(1-\frac{a^2}{b^2}\right)=a^2\left(e^2-1\right)=a^2\left(1-\frac{b^2}{a^2}-1\right) \\ & =-b^2 \\ & \Rightarrow \frac{y^2\left(b^2-a^2\right)}{b^2}=-b^2 \Rightarrow y^2=\frac{b^4}{\left(a^2-b^2\right)} \\ & \text { Height }=|y|=\frac{b^2}{\sqrt{a^2-b^2}} \end{aligned}$$
$$\begin{aligned} &\begin{aligned} & \text { Area }=(2 a e) \times \frac{1}{2} \times \frac{b^2}{\sqrt{a^2-b^2}}=30 \\ & =\frac{a b^2 e}{a \sqrt{1-\frac{b^2}{a^2}}}=b^2, a=\frac{17}{2} \end{aligned}\\ &\text { Distance between foci }=2 a e\\ &=17 \sqrt{1-\frac{b^2}{a^2}}=17 \sqrt{1-\frac{30 \times 4}{289}}=13 \end{aligned}$$
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