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The two common tangents to the parabola $y^{2} = 4 x$ and the ellipse $x^{2} + 4 y^{2} = 8$ meet at M. Of the two tangents, one meets the parabola and the ellipse at $P_{1}$ and $E_{1}$ respectively, and the other meets them at $P_{2}$ and $E_{2}$ respectively. Then, the area of the quadrilateral $P_{1} E_{1} E_{2} P_{2}$ is ___

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answer is 30.

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Detailed Solution

The point M is found to be $(- 4, 0)$ the equation of $E_{1} E_{2}$ is $x = - 2$ the equation of $P_{1} P_{2}$ is $x = 4$
$E_{1} = (- 2, 1), P_{1} = (4, 4), P_{2} = (4, - 4), E_{2} = (- 2, - 1)$
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$P_{1} E_{2} = \sqrt{6^{2} + 5^{2}} = \sqrt{61}$ ; $\frac{M E_{1}}{M P_{1}} = \frac{\sqrt{2^{2} + 1^{2}}}{\sqrt{8^{2} + 4^{2}}} = 1 / 4$
The area of $P_{1} E_{1} E_{2} P_{2} = \frac{1}{2} \times 8 \times 8 - 1 / 2 \times 2 \times 2 = 30$

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