Question

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

• A. $\left(\frac{a}{3},-\frac{2a}{\sqrt{3}}\right)$
• B. $(3a,-2\sqrt{3a})$
• C. $(3a,2\sqrt{3a})$
• D. $\left(\frac{a}{3},\frac{2a}{\sqrt{3}}\right)$

Answer 1

Answer: (A), (D)

$\left(\frac{a}{3},-\frac{2a}{\sqrt{3}}\right)$
$\left(\frac{a}{3},\frac{2a}{\sqrt{3}}\right)$

The point of contact with the tangent is given by $(\frac{a}{m^2},-\frac{2a}{m})$
where, $m=\tan (\pm \frac{\pi }{3})=\pm \sqrt{3}$
Therefore, point of contact is $(\frac{a}{3},\pm \frac{2a}{\sqrt{3}})$