The centre of a circle C is at the centre of the ellipse , a > b. Let C pass through the foci and 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 is 30 and the lengt

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Solution

Q.
The centre of a circle C is at the centre of the ellipse $E : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , a > 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 $P F_{1} F_{2}$ is 30 and the length of the major axis of E is 17, then the distance between the foci of E is : [2025]

1 26

2 13

3 12

4 $\frac{13}{2}$

Ans.
(2)

We have, the ellipse $(E) : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , (a > b)

$\Rightarrow x^{2} + \frac{a^{2} y^{2}}{b^{2}} = a^{2}$ ... (i)

The circle is $x^{2} + y^{2} = a^{2} e^{2}$ ... (ii)

Using (i) and (ii),

$\Rightarrow y^{2} (1 - \frac{a^{2}}{b^{2}}) = a^{2} (e^{2} - 1) = a^{2} (1 - \frac{b^{2}}{a^{2}} - 1) = - b^{2}$

$\Rightarrow y^{2} \frac{(b^{2} - a^{2})}{b^{2}} = - b^{2} \Rightarrow y^{2} \frac{b^{4}}{a^{2} - b^{2}} \Rightarrow | y | = \frac{b^{2}}{\sqrt{a^{2} - b^{2}}}$

Area of triangle $= \frac{1}{2} \times 2 a e \times \frac{b^{2}}{\sqrt{a^{2} - b^{2}}} = 30$

$\Rightarrow \frac{a b^{2} e}{a \sqrt{1 - \frac{b^{2}}{a^{2}}}} = b^{2} = 30$

Given, $2 a = 17 \Rightarrow a = \frac{17}{2}$ .

$∴$ Distance between the foci = 2ae

$= 17 \sqrt{1 - \frac{b^{2}}{a^{2}}} = 17 \sqrt{1 - \frac{30 \times 4}{289}} = 13$ .

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