Question

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

• A. $8$
• B. $6$
• C. $50$
• D. $32$
• E. $10$

Answer 1

The correct option is B

$6$

Explanation for the correct answer:

The equation of the ellipse $16{\left(x-2\right)}^2+25{\left(y+3\right)}^2=400$ can be written in standard form as

$\frac{X^2}{25}+\frac{Y^2}{16}=1$ $...\left(i\right)$

where, $X=x-2$ and $Y=y+3$

Comparing equation $\left(i\right)$ with $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ where $a>b$

we get $a^2=25$ and $b^2=16$

The eccentricity of the ellipse is given as

$e=\sqrt{1-\frac{b^2}{a^2}}$

$\Rightarrow$ $e=\sqrt{1-\frac{16}{25}}$

$\Rightarrow$ $e=\sqrt{\frac{9}{25}}$

$\Rightarrow$ $e=\frac{3}{5}$

Foci of the ellipse are given as $\left(X,Y\right)=\left(\pm ae,0\right)$

$\Rightarrow$ $x-2=\pm 5\times \frac{3}{5}$ and $y+3=0$

$\Rightarrow$ $x=5$ or $x=-1$ and $y=-3$

Hence, $\left(5,-3\right)$ and $\left(-1,-3\right)$ are the foci of the ellipse.

Distance between $\left(5,-3\right)$ and $\left(2,-3\right)$ $=\sqrt{{\left(5-2\right)}^2+{\left(-3+3\right)}^2}$

$=3$

Distance between $\left(-1,-3\right)$and $\left(2,-3\right)$$=\sqrt{{\left(-1-2\right)}^2+{\left(-3+3\right)}^2}$

$=3$

Hence, the sum of the distance between the foci of the given ellipse and the point $\left(2,-3\right)$ is $3+3=6$ units.

Hence, the option B is the correct option.