Q92 / 150 · MHT CET 2008 · Mathematics
The equation of common tangent to the circle $x^{2}+y^{2}=2$ and parabola $y^{2}=8 x$ is
- A$y=x+1$
- Correct answerB$y=x+2$
- C$y=x-2$
- D$y=-x+2$
Solution
Given, $y^{2}=8 x$
$$
\therefore \quad 4 a=8 \Rightarrow a=2
$$
Any tangent of parabola is,
$$
\begin{aligned}
y &=m x+\frac{a}{m} \\
\Rightarrow \quad & m x-y+\frac{2}{m}=0
\end{aligned}
$$
If it is a tangent to the circle $x^{2}+y^{2}=2$, then perpendicular from centre $(0,0)$ is equal to radius $\sqrt{2}$.
$$
\begin{array}{lr}
\therefore & \frac{\frac{2}{m}}{\sqrt{m^{2}+1}}=\sqrt{2} \\
\Rightarrow & \frac{4}{m^{2}}=2\left(m^{2}+1\right) \\
\Rightarrow & \quad m^{4}+m^{2}-2=0 \\
\Rightarrow & \left(m^{2}+2\right)\left(m^{2}-1\right)=0
\end{array}
$$
$\Rightarrow$
$m=\pm 1$
Hence, the common tangents are $y=\pm(x+2)$.
Hence, the correct answer is option: 'B'.