Question

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

• 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)

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

If the line $y=mx+c$ is a normal to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, then
$c^2=\frac{m^2(a^2-b^2{)}^2}{a^2+b^2{m}^2}$
[Here, $m=2,a^2=9$ and $b^2=16]$
$=\frac{\left(2{)}^2\right(9-16{)}^2}{9+16\times (2{)}^2}$
$=\frac{4\times 49}{9+64}=\frac{4\times 49}{73}=\frac{196}{73}$
$\therefore c=\frac{14}{\sqrt{73}}$