Question

If a tangent to a parabola $y^2=4ax$ makes an angle of $\frac{\pi }{3}$ with the axis of the parabola. Then point of contact(s) is/are

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

Answer 1

Answer: (A), (D)

The correct options are
A $(\frac{a}{3},-\frac{2a}{\sqrt{3}})$
D $(\frac{a}{3},\frac{2a}{\sqrt{3}})$

The axis of the parabola $y^2=4ax$ is $x-$axis
any line which makes an angle of $\frac{\pi }{3}$ will have slope as $m=\pm \tan \frac{\pi }{3}=\pm \sqrt{3}$
$\therefore y=\pm \sqrt{3}x\pm \frac{a}{\sqrt{3}}$
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 $y=\pm \frac{2a}{\sqrt{3}}$