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 $c^{2} = 8\left(4\right)^{2}+4 = 132$
$\Rightarrow c = \pm\sqrt{132}$