Question

The equations of the tangents to the ellipse $x^2+16y^2=16,$ each one of which makes an angle of ${60}^{\circ }$ with the $x-$axis, is

• A. $y=\sqrt{3}x\pm 1$
• B. $y=\sqrt{3}x\pm 3$
• C. $y=\sqrt{3}x\pm 5$
• D. $y=\sqrt{3}x\pm 7$

Answer 1

The correct option is D $y=\sqrt{3}x\pm 7$

We have, $x^2+16y^2=16$
$\Rightarrow \frac{x^2}{4^2}+\frac{y^2}{1^2}=1$
This is of the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$ where $a^2=16$ and $b^2=1$
So, the equation of the tangents with slope $m$ are
$y=mx\pm \sqrt{a^2{m}^2+b^2}\cdots \left(1\right)$
Given, $m=\tan {60}^{\circ }$
$\Rightarrow m=\sqrt{3}$
From equation $\left(1\right)$, we get
$y=\sqrt{3}x\pm \sqrt{16\times 3+1}$
$\Rightarrow y=\sqrt{3}x\pm 7$