Question

Let S and T be the foci of the ellipse $\frac{x^2}{16}+\frac{y^2}{8}=1$. If P(x,y) is any point on the ellipse, then the maximum area of the triangle PST (in square units) is ___

• A. $2\sqrt{2}$
• B. $4\sqrt{2}$
• C. 8
• D. $8\sqrt{2}$

Answer 1

The correct option is C

8

The given ellipse is
$\frac{x^2}{16}+\frac{y^2}{8}=1a^2=16;b^2=8\text{Focal length, c =}\sqrt{16-8}=\sqrt{8}$

The foci of the ellipse lie on the major axis. The greatest perpendicular distance from the major axis to any point on the ellipse is the length of the semi-minor axis.

For the triangle PST,
$ST=2S=2\sqrt{8}\text{Max Area =}\frac{1}{2}\times ST\times b=\frac{1}{2}\times 2\sqrt{8}\times \sqrt{8}=\text{8 sq. units}$