Ellipse Ques 3
- Let $P$ be a variable point on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ 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… .
(1994, 2M)
Show Answer
Answer:
Correct Answer: 3.$b\sqrt a^2 - b^2$
Solution:
- Given,
$ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 $
Foci $F _1$ and $F _2$ are $(-a e, 0)$ and $(a e, 0)$, respectively. Let $P(x, y)$ be any variable point on the ellipse.
The area $A$ of the triangle $P F _1 F _2$ is given by
$ \begin{alignedat} A & =\frac{1}{2}\left|\begin{array}{ccc} x & y & 1 \\ -a e & 0 & 1 \\ a e & 0 & 1 \end{array}\right| \\ & =\frac{1}{2}(-y)(-a e \times 1 + a e \times 1) \end{aligned} $
$ =-\frac{1}{2} y(-2 a e)=a \text { ey }=a e \cdot b \sqrt{1-\frac{x^{2}}{a^{2}}} $
So, $A$ is maximum when $x=0$.
$\therefore$ Maximum of $A=a b e=a b \sqrt{1-\frac{b^{2}}{a^{2}}}=a b \sqrt{\frac{a^{2}-b^{2}}{a^{2}}}$
$ =b \sqrt{a^{2}-b^{2}} $
BETA
