If $S$ and $S^{\prime}$ are the foci of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ and if $P S P^{\prime}$ is a focal chord with $S P=8$, then $S S^{\prime}$ is equal to

• Correct answer
  A

  $4+S^{\prime} P$
• B

  $S^{\prime} P-1$
• C

  $4+S P$
• D

  $S P-1$

Solution

Given equation of ellipse $$ \frac{x^2}{25}+\frac{y^2}{16}=1 $$ Here, $a=5$ and $b=4$ $$ \begin{aligned} e & =\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{16}{25}} \\ & =\sqrt{\frac{9}{25}}=\frac{3}{5} \end{aligned} $$
Also, $\quad P S+P S^{\prime}=2 \mathrm{a}=2 \times 5=10$ $$ \begin{array}{lll} \Rightarrow & P S^{\prime}=10-8 & {[\because S P=8]} \\ \Rightarrow & P S^{\prime}=2 & \end{array} $$ Now, $\quad S S^{\prime}=2 \mathrm{ae}=2 \times 5 \times \frac{3}{5}=6$ From options, $$ S S^{\prime}=4+S^{\prime} P $$

Hence, the correct answer is option: 'A'.