Let a focus of the ellipse E:x^2a^2+y^2b^2=1 be S(4,0) and its eccentricity be (4)/(5).

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 $	rianglePOS$ is equal to:

Accepted Answer

For the ellipse

$E:\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1$

a focus is $S(4,0)$ and eccentricity is

$e=\frac{4}{5}$

We need the area of $	rianglePOS$, where $P(3,\alpha )$ lies on the ellipse and $O=(0,0)$.

Since the focus is on the $x$-axis, the ellipse has major axis along the $x$-axis.

For such an ellipse, focus is at $(\pm ae,0)$.

Given one focus is $(4,0)$, so

$ae=4$

and $e=\frac{4}{5}$. Therefore,

$a\cdot \frac{4}{5}=4$

$a=5$

Now,

$c=ae=4$

and for ellipse,

${b}^{2}={a}^{2}-{c}^{2}$

So,

${b}^{2}=25-16=9$

Hence the ellipse is

$\frac{{x}^{2}}{25}+\frac{{y}^{2}}{9}=1$

Since $P(3,\alpha )$ lies on the ellipse, substitute $x=3$:

$\frac{{3}^{2}}{25}+\frac{{\alpha }^{2}}{9}=1$

$\frac{9}{25}+\frac{{\alpha }^{2}}{9}=1$

$\frac{{\alpha }^{2}}{9}=1-\frac{9}{25}=\frac{16}{25}$

${\alpha }^{2}=\frac{144}{25}$

$\alpha =\pm \frac{12}{5}$

Now find the area of triangle with vertices

$O(0,0),\, P(3,\alpha ),\, S(4,0)$

Using base $OS$:

$OS=4$

Since $OS$ lies on the $x$-axis, the height from $P$ to $OS$ is $|\alpha |$.

So area is

$\frac{1}{2}	imes 4	imes |\alpha |=2|\alpha |$

$=2	imes \frac{12}{5}=\frac{24}{5}$

Therefore, the area of $	rianglePOS$ is

$\frac{24}{5}$

So, the correct option is:

Solutions · 1