Question

The line $y = 2x + c$ touches the ellipse $\frac{x^2}{16}+\frac{y^2}{4}=1$ if c is equal to

• A. 0
• B. $\pm 2\sqrt{17}$
• C. $c=\pm \sqrt{15}$
• D. $c=\pm \sqrt{17}$

Answer 1

Answer: (B)

$\pm 2\sqrt{17}$

Answer 2

$y= 2x+c$ touches $\frac{x^{2}}{16}+\frac{y^{2}}{4} =1$ if
$c^{2} = a^{2}m^{2}+b^{2}$
$\quad\left({\text{Here}}\,\, a^{2}=16, b^{2}=4, m=2\right)$
i.e.,if $c^{2}=16\left(4\right)+4=68$
$\therefore c = \pm2\sqrt{17}$