Question

If line $y=2x+c$ 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)

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 $b^2=16$]
$=\frac{{\left(2\right)}^2{\left(9−16\right)}^2}{9+16×{\left(2\right)}^2}$
$=\frac{4×49}{9+64}=\frac{4×49}{73}=\frac{196}{73}$
$∴c=\frac{14}{\sqrt{73}}$