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Let S and S 1...

Question

Let $S$ and $S_{1}$ be the foci of the ellipse $9 x^{2} + 5 y^{2} = 30 y$ and a point $P$ be $(3, 3) .$ The area of triangle $P S S_{1}$ is

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Solution

Given ellipse, $9 x^{2} + 5 y^{2} = 30 y$ can be written as
$9 x^{2} + 5 (y^{2} - 6 y) = 0$
$= 9 x^{2} + 5 (y - 3)^{2} = 45$
$= \frac{x^{2}}{5} + \frac{(y - 3)^{2}}{9} 1$
Centre is (0, 3)
Foci are $S (0, 3 + \sqrt{9 - 5}), S_{1} (0, 3 - \sqrt{9 - 5})$
$\Rightarrow S (0, 5), S_{1} (0, 1)$
Area of $Δ P S S_{1} = \frac{1}{2} ∣ ∣ ∣ ∣ \begin{matrix} 3 & 3 & 1 0 & 5 & 1 0 & 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ = \frac{12}{2} = 6$

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