Let a focus of the ellipse E:x^2a^2+y^2b^2=1 be S(4,0) and its eccentricity be 45.

Question

Let a focus of the ellipse $E:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ be $S(4,0)$ and its eccentricity be $\dfrac45$. If the point $P(3,\alpha)$ lies on $E$ and $O$ is the origin, then the area of $	riangle POS$ is equal to:

Accepted Answer

Since focus is on the $x$-axis, major axis is along $x$: $ae=4$ with $e=\dfrac45\Rightarrow a=5,\ c=4$.

$$b^2=a^2-c^2=25-16=9 \;\Rightarrow\; E:\frac{x^2}{25}+\frac{y^2}{9}=1.$$

At $x=3$: $\dfrac{9}{25}+\dfrac{\alpha^2}{9}=1\Rightarrow \alpha^2=\dfrac{144}{25}\Rightarrow \alpha=\dfrac{12}{5}$.

With base $OS=4$ on the $x$-axis and height $|\alpha|=\dfrac{12}{5}$:

$$\text{Area}=\frac12\times4\times\frac{12}{5}=\frac{24}{5}.$$

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