Question

The sum of the distances of a point $(2,-3)$ from the foci of an ellipse $16{(x-2)}^2+25(y^4-$ $3{)}^2=400$ is

• A. 8
• B. 6
• C. 50
• D. 32
• E. 10

Answer 1

Answer: (B)

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, $a>b$
$\therefore$ $e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{16}{25}}=\frac{3}{5}$
$\therefore$ Focus $(\pm ae,0)=(\pm 3,0)$
$\Rightarrow$ $x-2=\pm 3,y+3=0$
$\Rightarrow$ $x=5,-1,y=-3$
$\therefore$ 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
$=3+3=6$