Question

The equation of the common tangent with positive slope to the parabola $y^2=8\sqrt{3}x$ and the hyperbola $4x^2-y^2=4$ is:

• A. $y=\sqrt{6}x+\sqrt{2}$
• B. $y=\sqrt{6}x-\sqrt{2}$
• C. $y=\sqrt{3}x+\sqrt{2}$
• D. $y=\sqrt{3}x-\sqrt{2}$

Answer 1

The correct option is A $y=\sqrt{6}x+\sqrt{2}$

Equation of tangent in slope form of parabola $y^2=8\sqrt{3}x$ is $y=mx+c$ ......(i)
where, $c=\frac{a}{m}$
$\therefore c=\frac{2\sqrt{3}}{m}$ .....(ii)
Also, tangent to the hyperbola
$4x^2-y^2=4$
or $\frac{x^2}{1}-\frac{y^2}{4}=1$ is
$c^2=a^2{m}^2-b^2$
$c^2=1m^2-4$
$\Rightarrow {\left(\frac{2\sqrt{3}}{m}\right)}^2=m^2-4$ [from equation (ii)]
$\Rightarrow \frac{12}{m^2}=m^2-4$
$\Rightarrow m^4-4m^2-12=0$
$\Rightarrow m^4-6m^2+2m^2-12=0$
$\Rightarrow m^2(m^2-6)+2(m^2-6)=0$
$\Rightarrow (m^2+2)(m^2-6)=0$
$\Rightarrow m^2-6=0$ and $m^2+2\ne 0$
$\Rightarrow m^2=6$
$\Rightarrow m=\pm \sqrt{6}$
i.e., $m=\sqrt{6}$ as $m$ is positive slope.
$\therefore$ from equation (i),
$y=\sqrt{6}x+\frac{2\sqrt{3}}{\sqrt{6}}$
$\Rightarrow y=\sqrt{6}x+\sqrt{2}$