Question

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

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

Answer 1

The correct option is C $\left(\frac{-15}{4},\frac{\sqrt{63}}{4}\right)$

We have ellipse as,

${9x}^2+25y^2=225$

$\frac{x^2}{5^2}+\frac{y^2}{3^2}=1$

$\therefore a=5$

$b=3$

$c=\sqrt{a^2-b^2}$

$c=4$

as sum of distance of both focus remains same,

distance from $F_1$, of the point is 2.

$\sqrt{{[x-(-4\left)\right]}^2+y^2}=2$

$\sqrt{{(-x+4)}^2+y^2}=2$

${(4-x)}^2+y^2=4⟶\left(1\right)$

distance from $F_2$, of the point is $4\times 2=8$.

$\sqrt{{(-x-4)}^2+y^2}=8$

${(x+4)}^2+y^2=64⟶\left(2\right)$

$\left(2\right)-\left(1\right)$,

We get,

${(x+4)}^2-{(x-4)}^2=60$

$-16x=60$

$x=\frac{-15}{4}$

put x in (1),

${(4-\frac{15}{4})}^2+y^2=4$

${\left(\frac{1}{4}\right)}^2+y^2=4$

$y^2=4-\frac{1}{16}$

$y^2=\frac{64-1}{16}$

$\therefore y=\frac{\sqrt{63}}{4}$