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)

4

Answer 2

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
$\sqrt{\left(4e-0\right)^{2}+\left(0-3\right)^{2}} = \sqrt{16e^{2}+9}$
$\sqrt{\frac{16\times7}{16}+9} = \sqrt{7+9}$
$= \sqrt{16} = 4$