Question

A tangent to the parabola $y^2=4ax$ makes an angle of ${60}^o$ with the axis ; find its point of contact.

Answer 1

$\theta ={60}^{\circ }m=\tan \theta \Rightarrow m=\tan {60}^{\circ }=\sqrt{3}$

Equation of tangent is

$y=mx+\frac{a}{m}y=\sqrt{3}x+\frac{a}{\sqrt{3}}\sqrt{3}y=3x+a\Rightarrow x=\frac{\sqrt{3}y-a}{3}$

Susbstituting $x$ in equation of parabola

$y^2=4ax{y}^2=4a\left(\frac{\sqrt{3}y-a}{3}\right)3y^2=4a\sqrt{3}y-4a^{2}3y^2-4a\sqrt{3}y+4a^2=0{(\sqrt{3}y-2a)}^2=0\Rightarrow y=\frac{2a}{\sqrt{3}}x=\frac{\sqrt{3}y-a}{3}\Rightarrow x=\frac{\sqrt{3}\frac{2a}{\sqrt{3}}-a}{3}=\frac{a}{3}$

So, the point of contact is $(\frac{a}{3},\frac{2a}{\sqrt{3}})$