Question

The distance of a focus of the ellipse $9x^2+16y^2=144$ from an end of the minor axis is

• A. $\frac{3}{2}$
• B. 3
• C. 4
• D. None of these

Answer 1

Answer: (C)

Ellipse is $\frac{x^2}{16}+\frac{y^2}{9}=1$
$a^2=16,b^2=9$. Focus is $\left(ae,0\right)$ i.e., $\left(4e,0\right)$
$b^2=a^2\left(1-e^2\right)$ gives $\frac{9}{16}=1-e^2$
$\Rightarrow e^2=1-\frac{9}{16}=\frac{7}{16}$
$\Rightarrow e=\frac{\sqrt{7}}{4}$
One end of mirror axis is $B\left(0,3\right)$
Distance of the focus from this end
$\sqrt{{\left(4e-0\right)}^2+{\left(0-3\right)}^2}=\sqrt{16e^2+9}$
$\sqrt{\frac{16\times 7}{16}+9}=\sqrt{7+9}$
$=\sqrt{16}=4$