JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
On the ellipse \(\frac{x^{2}}{8}+\frac{y^{2}}{4}=1\) let \(P\) be a point in the second quadrant such that the tangent at \(\mathrm{P}\) to the ellipse is perpendicular to the line \(x+2 y=0\). Let \(S\) and \(\mathrm{S}^{\prime}\) be the foci of the ellipse and \(\mathrm{e}\) be its eccentricity. If \(\mathrm{A}\) is the area of the triangle \(SPS'\) then, the value of \(\left(5-\mathrm{e}^{2}\right) . \mathrm{A}\) is :
- A \(12\)
- B \(6\)
- C \(14\)
- D \(24\)
Answer & Solution
Correct Answer
(B) \(6\)
Step-by-step Solution
Detailed explanation
Equation of tangent : \(\mathrm{y}=2 \mathrm{x}+6\) at \(\mathrm{P}\) \(\therefore \mathrm{P}(-8 / 3,2 / 3)\) \(\mathrm{e}=\frac{1}{\sqrt{2}}\) \(\mathrm{~S} \& \mathrm{~S}^{\prime}=(-2,0) \&(2,0)\) Area of \(\Delta\) SPS' \(=\frac{1}{2} \times 4 \times \frac{2}{3}\)…
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