Question

The equation of common tangent to the circle $x^2+y^2=2$ and parabola $y^2=8x$ is

• A. $y=x+1$
• B. $y=x+2$
• C. $y=x-2$
• D. $y=-x+2$

Answer 1

Answer: (B)

Given, $y^2=8x$
$\therefore 4a=8\Rightarrow a=2$
Any tangent of parabola is,
$y=mx+\frac{a}{m}$
$\Rightarrow mx-y+\frac{2}{m}=0$
If it is a tangent to the circle $x^2+y^2=2,$ then perpendicular from centre (0,0) is equal to radius $\sqrt{2}$
$\therefore \frac{\frac{2}{m}}{\sqrt{m^2+1}}=\sqrt{2}$
$\Rightarrow \frac{4}{m^2}=2\left(m^2+1\right)$
$\Rightarrow m^4+m^2-2=0$
$\Rightarrow \left(m^2+2\right)\left(m^2-1\right)=0$
$\Rightarrow m=\pm 1$
Hence, the common tangents are $y=\pm (x+2)$