Q84 / 200 · WBJEE 2014 · Mathematics
The equation of the common tangent with positive slope to the parabola $y^{2}=8 \sqrt{3} x$ and the hyperbola $4 x^{2}-y^{2}=4$ is
- Correct answerA$y=\sqrt{6} x+\sqrt{2}$
- B$y=\sqrt{6} x-\sqrt{2}$
- C$y=\sqrt{3} x+\sqrt{2}$
- D$y=\sqrt{3} x-\sqrt{2}$
Solution
Equation of tangent in slope form of parabola $y^{2}=8 \sqrt{3} x$ is
$$
y=m x+c
$$
where, $\quad c=\frac{a}{m}$
$\therefore$
$c=\frac{2 \sqrt{3}}{m}$
Also, tangent to the hyperbola
$$
4 x^{2}-y^{2}=4
$$
or $\frac{x^{2}}{1}-\frac{y^{2}}{4}=1$ is
$$
\begin{array}{l}
c^{2}=a^{2} m^{2}-b^{2} \\
c^{2}=1 m^{2}-4
\end{array}
$$
$\Rightarrow \quad\left(\frac{2 \sqrt{3}}{m}\right)^{2}=m^{2}-4$
$\Rightarrow \quad \frac{12}{m^{2}}=m^{2}-4$
$\Rightarrow \quad m^{4}-4 m^{2}-12=0$
$\Rightarrow \quad m^{4}-6 m^{2}+2 m^{2}-12=0$
$\Rightarrow \quad m^{2}\left(m^{2}-6\right)+2\left(m^{2}-6\right)=0$
$\Rightarrow \quad\left(m^{2}+2\right)\left(m^{2}-6\right)=0$
$\Rightarrow \quad m^{2}-6=0$ and $m^{2}+2 \neq 0$
$\Rightarrow \quad m^{2}=6$
$\Rightarrow \quad m=\pm \sqrt{6}$
i.e $m=\sqrt{6}$ as $m$ is positive slope.
$\therefore$
$y=\sqrt{6} x+\frac{2 \sqrt{3}}{\sqrt{6}}$
$\Rightarrow \quad y=\sqrt{6} x+\sqrt{2}$
Hence, the correct answer is option: 'A'.