Q52 / 120 · VITEEE 2016 · Mathematics
The equation of one of the common tangents to the parabola $y^{2}=8 x$ and $x^{2}+y^{2}-12 x+4=0$ is
- A$y=-x+2$
- B$\mathrm{y}=\mathrm{x}-2$
- Correct answerC$y=x+2$
- DNone of these
Solution
Any tangent to parabola $y^{2}=8 x$ is $y=$
$$
\mathrm{mx}+\frac{2}{\mathrm{~m}}......(i)
$$
It touches the circle $x^{2}+y^{2}-12 x+4=0$ if the length of perpendicular from the centre $(6,0)$ is equal to radius $\sqrt{32}$.
$\begin{array}{l}
\therefore \frac{6 \mathrm{~m}+\frac{2}{\mathrm{~m}}}{\sqrt{\mathrm{m}^{2}+1}}=\pm \sqrt{32} \\
\Rightarrow\left(3 \mathrm{~m}+\frac{1}{\mathrm{~m}}\right)^{2}=8\left(\mathrm{~m}^{2}+1\right) \\
\Rightarrow\left(3 \mathrm{~m}^{2}+1\right)^{2}=8\left(\mathrm{~m}^{4}+\mathrm{m}^{2}\right) \\
\Rightarrow \mathrm{m}^{4}-2 \mathrm{~m}^{2}+1=0 \Rightarrow \mathrm{m}=\pm 1
\end{array}$
Hence, the required tangents are $y=x+2$ and $y=-x-2$.
Hence, the correct answer is option: 'C'.