Question

If the straight line $y=4x+c$ is a tangent to the ellipse $\frac{x^2}{8}+\frac{y^2}{4}=1$ then $c$ will be equal to :

• A. $\pm$4
• B. $\pm$ 6
• C. $\pm$1
• D. $\pm \sqrt{(132)}$

Answer 1

Answer: (D)

$y=mx+c$ touches $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$if
$c^2=a^2{m}^2+b^2$
$\therefore y=4x+c$ touches $\frac{x^2}{8}+\frac{y^2}{4}=1$
if $c^2=8{\left(4\right)}^2+4=132$
$\Rightarrow c=\pm \sqrt{132}$