Question

On the ellipse $9x^2+25y^2=225,$ then find the point the distance from which to our focus $F_1$ is four times the distance to the other focus $F_2$

• A. $\left(\frac{-15}{4},\frac{\sqrt{63}}{2}\right)$
• B. $\left(\frac{-15}{4},\frac{\sqrt{63}}{4}\right)$
• C. $\left(\frac{-15}{2},\frac{\sqrt{63}}{2}\right)$
• D. $(-15,\sqrt{63})$

Answer 1

The correct option is B

$\left(\frac{-15}{4},\frac{\sqrt{63}}{4}\right)$

Explanation for the correct option.

Finding On the ellipse $9x^2+25y^2=225,$ then find the point the distance from which to our focus $F_1$ is four times the distance to the other focus $F_2$

Given: Ellipse $9x^2+25y^2=225,$
$$\begin{gathered} \frac{x^2}{5^2}+\frac{y^2}{3^2}=1 \\ \therefore a=5 \\ b=3 \\ c=\sqrt{a^2-b^2} \\ c=4 \end{gathered}$$

as sum of distance of both focus remains same, distance from $F_1$, of the point is $2.$
$$\begin{gathered} \sqrt{{[x-(-4\left)\right]}^2+y^2}=2 \\ \sqrt{{(-x+4)}^2+y^2}=2 \\ {(4-x)}^2+y^2=4....\left(1\right) \end{gathered}$$

distance from $F_2$, of the point is $4\times 2=8.$
$$\begin{gathered} \sqrt{{(-x-4)}^2+y^2}=8 \\ {(x+4)}^2+y^2=64....\left(2\right) \end{gathered}$$

Subtracting $2$ from $1$ ,We get;

$$\begin{gathered} {(x+4)}^2-{(x-4)}^2=60 \\ -16x=60 \\ x=\frac{-15}{4} \end{gathered}$$

put $x$ in $\left(1\right)$,

$$\begin{gathered} {\left(4-\frac{15}{4}\right)}^2+y^2=4 \\ {\left(\frac{1}{4}\right)}^2+y^2=4 \\ {y}^2=4-\frac{1}{16} \\ {y}^2=\frac{64-1}{16} \\ \therefore y=\frac{\sqrt{63}}{4} \end{gathered}$$

Hence the correct answer is option (B)