Question

$S$ and $S^{{}^{\prime }}$ are the foci of the ellipse $25x^2+16y^2=1600$, then the sum of the distances from $S$ and $S^{\prime }$ to the point $(4\sqrt{3},5)$ is:

• A. $20$
• B. $15$
• C. $40$
• D. $30$

Answer 1

The correct option is A $20$

Given ellipse is $25x^2+16y^2=1600$

This can be written as $\frac{x^2}{64}+\frac{y^2}{100}=1$

Compare it to standard form of ellipse to get $a=8,b=10$

here $a<b$

So eccentricity e$=\sqrt{1-\frac{a^2}{b^2}}=\sqrt{1-\frac{64}{100}}=\frac{3}{5}$

Now the foci are given by $(0,\pm \sqrt{b^2-a^2})$

So $foci:(0,\pm \sqrt{36})$

$S:(0,6),S_1:(0,-6)$

Now find distance of $S$ and $S_1$ from the given point $(4\sqrt{3},5)$

So sum$=\sqrt{{\left(4\sqrt{3}\right)}^2+1}+\sqrt{{\left(4\sqrt{3}\right)}^2+{\left(11\right)}^2}=7+13=20$