If $P$ be a point on ellipse $4 x^{2} + y^{2} = 8$ with eccentric angle $\frac{π}{4}$ Tangent at $P$ intersects the $X, Y$ axes at $A, B$ respectively, normal at $P$ intersects the $X, Y$ axes at $A', B'$ respectively. Then the ratio of area of $△ APA'$ and area of $△ {BPB}^{'}$ is

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3:1

1:1

4:1

2:1

answer is D.

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Detailed Solution

Clearly P = (1, 2)
Equation of tangent at P is $4 x (1) + y (2) = 8$
$\Rightarrow A = (2, 0) and B = (0, 4)$
Similarly normal at P is $2 x - 4 y = 6$
$\Rightarrow A^{'} (- 3, 0) and B^{'} (0, \frac{3}{2})$
$∴$ Area BPB $= \frac{1}{2} |\begin{array}{lll} 0 4 1 \\ 1 2 1 \\ 0 \frac{3}{2} 1 \end{array}| = \frac{5}{4} and Area A_{P A}^{'} = 5$
$∴$ Ratio=4