Question

The equation(s) of the common tangent(s) to the parabola $y^2=2x$ and the circle $x^2+y^2+4x=0$ is/are

• A. $2\sqrt{6}x+y=12$
• B. $x+2\sqrt{6}y+12=0$
• C. $x-2\sqrt{6}y+12=0$
• D. $2\sqrt{6}x-y=12$

Answer 1

Answer: (B), (C)

$x+2\sqrt{6}y+12=0$
$x-2\sqrt{6}y+12=0$

Let $y=mx+c$ be the equation of common tangent to the parabola $y^2=2x$ and the circle $x^2+y^2+4x=0$.
The condition for $y=mx+c$ to become tangent to $y^2=4ax$ is $c=\frac{a}{m}$
$\Rightarrow c=\frac{1}{2m}$ ----(1)
The line $y=mx+c$ to become tangent to the circle, the perpendicular distance from the centre to the line should be equal to radius of the circle.
i.e., $\frac{-2m+c}{\sqrt{1+m^2}}=2$ ----(2)
Solving (1) and (2) gives
$m=\pm \frac{1}{2\sqrt{6}}$
$\therefore$ Equations of common tangents are $x+2\sqrt{6y}+12=0$ and $x-2\sqrt{6y}+12=0$.
Hence, options B and C are the correct answers.