Question

The equation of the common tangent of the parabola $y^2=8x$ and $x^2+12y^2=48$ is

• A. $y=\frac{x}{\sqrt{3}}+2\sqrt{3}$
• B. $y=-\frac{x}{\sqrt{3}}-2\sqrt{3}$
• C. $y=\frac{x}{2}+4$
• D. $y=-\frac{x}{2}-4$

Answer 1

Answer: (C), (D)

The correct options are
C $y=\frac{x}{2}+4$
D $y=-\frac{x}{2}-4$

The tangent to $y^2=8x$ is $y=mx+\frac{2}{m}...\left(1\right)$
The tangent to $x^2+12y^2=48$ is $y=mx\pm \sqrt{48m^2+4}...\left(2\right)$
Comparing equation $\left(1\right)$ and $\left(2\right)$
$\frac{2}{m}=\pm \sqrt{48m^2+4}$
$\frac{4}{m^2}=48m^2+4$
$\Rightarrow 1=12m^4+m^2$
$\Rightarrow 12m^4+m^2-1=0$
$\Rightarrow (4m^2-1)(3m^2+1)=0$
$(3m^2+1)\ne 0,4m^2=1$
$m=\pm \frac{1}{2}$
The equations of the tangents are
$y=\frac{x}{2}+4$ and $y=-\frac{x}{2}-4$