Question

The equation of the tangent to the parabola $y^{2}=12 x$, which makes an angle $30^{\circ}$ with the positive direction of $x$ axis is given by $x-\sqrt{3} y+9=0$, then its point of contact is

• A. $(-9,-6 \sqrt{3})$
• B. $(9,-6 \sqrt{3})$
• C. $(-9,6 \sqrt{3})$
• D. $(9,6 \sqrt{3})$

Answer 1

Answer: (D)

Let the point of tangency of tangent
$x-\sqrt{3} y+9=0$ to the parabola $y^{2}=12 x$ is
$\left(x_{1} y_{1}\right)$
Since equation of tangent to the parabola
$y^{2}=12 x$ at point $\left(x_{1}, y_{1}\right)$ is
$y y_{1}=6\left(x +x_{1}\right)$
$6 x-y_{1} y+6 x_{1}=0$, which represent the tangent
$x-\sqrt{3} y+9=0$, so on comparing, we get
$\frac{6}{1}=\frac{-y_{1}}{-\sqrt{3}}=\frac{6 x_{1}}{9}$
$\Rightarrow \left(x_{1}, y_{1}\right)=(9,6 \sqrt{3})$
Therefore, point of contact is $(9,6 \sqrt{3})$.