Question

The equation of the tangent to the parabola $y^2=12x$, 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 $y^2=12x$ is
$\left(x_1{y}_1\right)$
Since equation of tangent to the parabola
$y^2=12x$ at point $\left(x_1,y_1\right)$ is
$y{y}_1=6\left(x+x_1\right)$
$6x-y_{1}y+6x_1=0$, which represent the tangent
$x-\sqrt{3}y+9=0$, so on comparing, we get
$\frac{6}{1}=\frac{-y_1}{-\sqrt{3}}=\frac{6x_1}{9}$
$\Rightarrow \left(x_1,y_1\right)=(9,6\sqrt{3})$
Therefore, point of contact is $(9,6\sqrt{3})$.