Question

The equation of one of the common tangents to the parabola $y^2=8x$ and $x^2+y^2-12x+4=0$ is

• A. $y=-x+2$
• B. $y=x-2$
• C. $y=x+2$
• D. None of these

Answer 1

Answer: (C)

Any tangent to parabola $y^2=8x$ is $y=mx+\frac{2}{m}\dots$(i)
It touches the circle $x^2+y^2-12x+4=0$,
if the length of perpendicular from the centre $(6,0)$ is equal to radius $\sqrt{32}$.
$\therefore \frac{6m+\frac{2}{m}}{\sqrt{m^2+1}}=\pm \sqrt{32}$
$\Rightarrow {\left(3m+\frac{1}{m}\right)}^2=8\left(m^2+1\right)$
$\Rightarrow {\left(3m^2+1\right)}^2=8\left(m^4+m^2\right)$
$\Rightarrow m^4-2m^2+1=0$
$\Rightarrow m=\pm 1$
Hence, the required tangents are $y=x+2$ and $y=-x-2$.