Question

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

• A. $8$
• B. $12$
• C. $16$
• D. $20$

Answer 1

The correct option is B

$12$

Explanation for the correct option:

Determining the maximum value of $∆$:

As seen in the equation, $\mathrm{a}>\mathrm{b}$,

So major axis is the x-axis

Foci $=\left(\pm \mathrm{ae},0\right)$

$$\begin{gathered} \mathrm{e}=\sqrt{1-\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}} \\ =\sqrt{1-\frac{16}{25}} \\ =\frac{3}{5} \end{gathered}$$

Foci $=\left(\pm 5\times \frac{3}{5},0\right)$

$=\left(\pm 3,0\right)$

Base of the triangle $=$ Distance between two focus

$=$$2\mathrm{ae}$

Area of triangle $∆\mathrm{PSS}\text{'}=\frac{1}{2}\times \mathrm{base}\times \mathrm{height}$

$=\frac{1}{2}\times 6\times \mathrm{height}$

For maximum area, height should be maximum

For maximum height, the point $\mathrm{P}$ should lie on the minor vertical point

So height is $\mathrm{b}=4$

therefore, Maximum area of $∆\mathrm{PSS}\text{'}=\frac{1}{2}\times \mathrm{base}\times \mathrm{height}$

$$\begin{gathered} =\frac{1}{2}\times 6\times 4 \\ =12\mathrm{square}\mathrm{units} \end{gathered}$$

Hence, option $\left(\mathrm{B}\right)$ is the correct answer.