The two common tangents to the parabola y2 = 4x and the ellipse x2 + 4y2 = 8 meet at M of the two tangents, one meets the parabola and the ellipse at P1 and E1respectively, and the other meets them at P2 and E2 respectively. Then the area of the quadrilateral P1E1E2P2 is

The point M is found to be(–4, 0) the equation of E1E2 is x = –2 the equation of P1P2 is x = 4E1=(−2,1),P1=(4,4),P2=(4,−4),E2=(−2,−1)P1E2=62+52=61;ME1MP1=22+1282+42=14The area ofP1E1E2P2=12×8×8−1/2×2×2=30

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The two common tangents to the parabola y² = 4x and the ellipse x² + 4y² = 8 meet at M of the two tangents, one meets the parabola and the ellipse at P₁ and E₁respectively, and the other meets them at P₂ and E₂ respectively. Then the area of the quadrilateral P₁E₁E₂P₂ 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 E₁E2 is x = –2

the equation of P₁P₂ is x = 4
$E_{1} = (- 2, 1), P_{1} = (4, 4), P_{2} = (4, - 4), E_{2} = (- 2, - 1)$

$P_{1} E_{2} = \sqrt{6^{2} + 5^{2}} = \sqrt{61}; \frac{{ME}_{1}}{{MP}_{1}} = \frac{\sqrt{2^{2} + 1^{2}}}{\sqrt{8^{2} + 4^{2}}} = \frac{1}{4}$
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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