Question

A tangent to a parabola $y^2=4ax$ is inclined at $\frac{\pi }{3}$ with axis of the parabola. The point of contact is

• A. $(\frac{a}{3},-\frac{2a}{\sqrt{3}})$
• B. $(3a,-2\sqrt{3}a)$
• C. $(3a,2\sqrt{3}a)$
• D. None of these

Answer 1

The correct option is A $(\frac{a}{3},-\frac{2a}{\sqrt{3}})$

Equation of the tangent to $y^2=4ax$ is $y=mx+\frac{a}{m}$

where $m=\tan \frac{\pi }{3}=\sqrt{3}$

$\therefore y=\sqrt{3}x+\frac{a}{\sqrt{3}}$

Point of contact of the tangent with the parabola can be obtained by

${(\sqrt{3}x+\frac{a}{\sqrt{3}})}^2=4ax\Rightarrow 3x^2+\frac{a^2}{3}+2ax=4ax$

$\Rightarrow 9x^2-6ax+a^2=0$ $\Rightarrow {(3x-a)}^2=0$

$\therefore x=\frac{a}{3}$ and hence $y=\pm \frac{2a}{\sqrt{3}}$