Question

If line $y = 2\,x + c$ is a normal to the ellipse $\frac{x^2}{9} + \frac{y^2}{16} = 1$, then

• A. $c = \frac{2}{3}$
• B. $c = \sqrt{ \frac{73}{5}}$
• C. $c = \frac{14}{\sqrt{73}}$
• D. $c = \sqrt{ \frac{5}{7}}$

Answer 1

Answer: (C)

$c = \frac{14}{\sqrt{73}}$

Answer 2

If the line $y = mx + c$ is a normal to the ellipse
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$, then
$c^{2} = \frac{m^{2}\left(a^{2} - b^{2}\right)^{2}}{a^{2} + b^{2} m^{2}}$
[Here, $m = 2 ,a^{2} = 9$ and $b^{2} = 16$]
$= \frac{\left(2\right)^{2} \left(9 - 16\right)^{2}}{9+16 \times\left(2\right)^{2}}$
$= \frac{4 \times49}{9+64} = \frac{4 \times49}{73}=\frac{196}{73}$
$\therefore \, \, c = \frac{14}{\sqrt{73}}$