The equation of the tangent to the ellipse $x^2+16 y^2=16$ which makes an angle $60^{\circ}$ with the…
The equation of the tangent to the ellipse $x^2+16 y^2=16$ which makes an angle $60^{\circ}$ with the $X$-axis is
$\sqrt{3} x-y+7=0$
) $\sqrt{3} x+y+7=0$
$\sqrt{3} x+y-7=0$
$\sqrt{3} x-y=0$
Solution
Equation of tangent to the ellipse $\frac{x^2}{16}+\frac{y^2}{1}=1$, have slope $m=\tan 60^{\circ}=\sqrt{3}$ is $\begin{aligned} & y=\sqrt{3} x \pm \sqrt{48+1} \\ \Rightarrow \quad & \sqrt{3} x-y+7=0 \text { or } \sqrt{3} x-y-7=0\end{aligned}$