Suppose, S and S' are foci of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$. If P is variable point on the ellipse and if $\Delta$ is area of the triangle PSS', then the maximum value of $\triangle$ is

Question

Suppose, $S$ and $S^{\prime}$ are foci of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$. If $P$ is variable point on the ellipse and if $\Delta$ is area of the triangle PSS', then the maximum value of $\triangle$ is

(a) 8

(b) 12

(c) 16

(d) 20

Answer 1

We have, $a^2=25$ and $b^2=16$

$\therefore \quad e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{16}{25}}=\sqrt{\frac{9}{25}}=\frac{3}{5}$

So, the coordinates of foci $S$ and $S^{\prime}$ are $(3,0)$ and $(-3,0)$ respectively. Let $P(5 \cos \theta, 4 \sin \theta)$ be a variable point on the ellipse.

Then, $\Delta=$ area of $\triangle P S S^{\prime}=\frac{1}{2} \times 6 \times 4 \sin \theta=12 \sin \theta$

So, maximum value of area of $\triangle P S S^{\prime}$ is 12 , since value of $\sin \theta$ lies between -1 and 1

So, The correct option of this question will be (B).