Let a focus of the ellipse E: frac{x^2}{a^2} | vedclass

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(C) Given the focus $S(ae, 0) = (4, 0)$ and eccentricity $e = 4/5$.
Thus, $ae = 4 \implies a(4/5) = 4 \implies a = 5$.
Using the relation $b^2 = a^2

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(C) Given the focus $S(ae, 0) = (4, 0)$ and eccentricity $e = 4/5$.
Thus, $ae = 4 \implies a(4/5) = 4 \implies a = 5$.
Using the relation $b^2 = a^2(1 - e^2)$, we get $b^2 = 25(1 - 16/25) = 25(9/25) = 9$.
The equation of the ellipse is $\frac{x^2}{25} + \frac{y^2}{9} = 1$.
Since point $P(3, \alpha)$ lies on the ellipse, we substitute $x=3$ and $y=\alpha$ into the equation:
$\frac{3^2}{25} + \frac{\alpha^2}{9} = 1 \implies \frac{9}{25} + \frac{\alpha^2}{9} = 1 \implies \frac{\alpha^2}{9} = 1 - \frac{9}{25} = \frac{16}{25}$.
$\alpha^2 = \frac{16 	imes 9}{25} \implies \alpha = \pm \frac{12}{5}$.
Taking $\alpha = 12/5$, the coordinates are $O(0, 0)$, $S(4, 0)$, and $P(3, 12/5)$.
The area of $	riangle POS = \frac{1}{2} |x_O(y_S - y_P) + x_S(y_P - y_O) + x_P(y_O - y_S)|$.
Area $= \frac{1}{2} |0(0 - 12/5) + 4(12/5 - 0) + 3(0 - 0)| = \frac{1}{2} |48/5| = 24/5$.

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 $\triangle POS$ is equal to: (in $/ 5$)

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