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27/12/2021 1:37 pm
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Ellipse x2/8 + y2/4 = 1 tangent at P(2nd quard) ⊥ to x + dy = 0 eccentricity = e. SS' is foci. Find (s - e2)*ΔSPS'.
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27/12/2021 1:45 pm
x2/8 + y2/4 = 1
e2 = 1 - \(\frac{4}{8}\) = \(\frac{1}{2}\)
Equation of tangent at P(x1, y1) with slope 2 is
2x - y + 6 = 0 .....(1)
Also, \(\frac{xx_1}{8}\) + \(\frac{yy_1}{4}\) = 1 .....(2)
From (1) and (2)
\(\frac{x_1}{16}\) = \(\frac{y_1}{-4}\) = \(\frac{-1}{6}\)
∴ (x1, y1) = \((\frac{-8}{3},\frac{2}{3})\)
∴ (5 - e2)*ΔSPS' = (5 - \(\frac{1}{2}\)).aey
= \(\frac{9}{2} \times 2 \times \frac{2}{3}\) = 6