Question
Let a focus of the ellipse $E:\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ be $S(4,0)$ and its eccentricity be $\frac{4}{5}$. If the point $P(3,\alpha)$ lies on E and O is the origin, then the area of $\bigtriangleup POS$ is equal to :
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Solution
Sol. [Key Concept: From $ae = 4$, $e = \frac{4}{5}$ get $a = 5$, $b^{2} = 9$; find $\alpha$ from the ellipse, then area $= \frac{1}{2}\ (OS)\ |\alpha|$.]
$$a > b $$Focus $(ae,0) = (4,0)$
$${\therefore ae = 4 }{a\left( \frac{4}{5} \right) = 4 \Rightarrow a = 5 }{\because b^{2} = a^{2}\left( 1 - e^{2} \right) }{b^{2} = 25\left( 1 - \frac{16}{25} \right) }{\therefore E:\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1 }$$∵ P lies on the ellipse E
$${\therefore\frac{9}{25} + \frac{\alpha^{2}}{9} = 1 }{\therefore\alpha = \pm \frac{12}{5} }$$∴ Area of $\Delta_{POS} = \frac{1}{2}(OS)(PN)$
$$= \frac{1}{2} \times 4 \times \frac{12}{5} = \frac{24}{5} $$
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