Question

The equation of the tangent to the parabola $y^2=16x$ inclined at an angle of ${60}^{\circ }$ to the positive $x-$axis is

• A. $3x-\sqrt{3}y+4=0$
• B. $3x+\sqrt{3}y+4=0$
• C. $3x-y+4=0$
• D. $3x+y+4=0$

Answer 1

The correct option is A $3x-\sqrt{3}y+4=0$

Given parabola $y^2=16x$
Inclination of the tangent is
$\theta ={60}^{\circ }\Rightarrow m=\tan {60}^{\circ }=\sqrt{3}$
Therefore equation of the tangent is of the form, $y=mx+\frac{a}{m}$
$\Rightarrow y=\sqrt{3}x+\frac{4}{\sqrt{3}}$
$\Rightarrow \sqrt{3}y=3x+4$