Question

Let P be a variable point on the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1.$ With foci $F_1$ and $F_2.$ If $A$ is the area of the triangle $P{F}_1{F}_2.$ then the maximum value of $A$ is

Answer 1

Given ellipse is, $\frac{x^2}{25}+\frac{y^2}{9}=1.$
$\Rightarrow a^2=25,b^2=9,\therefore e=\sqrt{1-\frac{9}{25}}=\frac{4}{5}$
$\Rightarrow F_1\equiv (-4,0),F_2\equiv (4,0)$
Let any point on the ellipse is, $P(5\cos \theta ,3\sin \theta )$
Thus area of triangle is $A=\frac{1}{2}|\left|\begin{array}{ccc}5\cos \theta & 3\sin \theta & 1\\ -4& 0& 1\\ 4& 0& 1\end{array}\right||=12\sin \theta$
We know maximum value of $\sin \theta$ is 1.
Hence $A_{max}=12$