JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
The centre of a circle C is at the centre of the ellipse \(E: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a \gt b\). 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 :
- A \(26\)
- B \(13\)
- C \(12\)
- D \(\frac{13}{2}\)
Answer & Solution
Correct Answer
(B) \(13\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \frac{1}{2} \mathrm{PF}_1 \cdot \mathrm{PF}_2=30 \\ & \mathrm{PF}_1+\mathrm{PF}_2=17 \\ & \mathrm{PF}_1=12 \mathrm{PF}_2=5 \\ & \mathrm{~F}_1 \mathrm{~F}_2=13 \\ & \text { option }(2)\end{aligned}\)
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