Question

The sum of the distances of a point $(2,-3)$ from the foci of an ellipse $16(x-2{)}^2+25(y+3{)}^2=400$ is

• A. 8
• B. 6
• C. 50
• D. 32

Answer 1

Answer: (B)

6

Given, equation of ellipse can be rewritten as

$\frac{(x-2{)}^2}{25}+\frac{(y+3{)}^2}{16}=1$

$\Rightarrow \frac{X^2}{25}+\frac{Y^2}{16}=1$

Where, $X=x-2,Y=y+3$

Here, $a>b$

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

Focus $(\pm ae,0)=(\pm 3,0)$

$\therefore x-2=\pm 3,y+3=0$

$\Rightarrow x=5,-1,y=-3$

Foci are $(-1,-3)$ and $(5,-3)$

Distance between $(2,-3)$ and $(-1,-3)$

$=\sqrt{(2+1{)}^2+(-3+3{)}^2}=3$

and distance between $(2,-3)$ and $(5,-3)$

$=\sqrt{(2-5{)}^2+(-3+3{)}^2}=3$

Hence, sum of the distance of point $(2,-3)$ from the foci $=3+3=6$