Question

The equation of the common tangent to the parabola $y^2=8x$ and the hyperbola $3x^2-y^2=3$ is

• A. $2x\pm y+1=0$
• B. $x\pm y+1=0$
• C. $x\pm 2y+1=0$
• D. $x\pm y+2=0$

Answer 1

The correct option is A $2x\pm y+1=0$

$3x^2-y^2=3$

$\Rightarrow$ $\frac{x^2}{1}-\frac{y^2}{3}=1$

Here, $a^2=1\Rightarrow a=1$

$b^2=3\Rightarrow b=\sqrt{3}$

Equation of tangent to hyperbola,

$y=mx\pm \sqrt{a^2{m}^2-b^2}$

$\Rightarrow$ $y=mx\pm \sqrt{m^2-3}$ ---- ( 1 )

Now,

$y^2=8x$

$4a=8$

$\Rightarrow$ $a=2$

Equation of tangent to parabola,

$y=mx+\frac{a}{m}$

$\Rightarrow$ $y=mx+\frac{2}{m}$ ----- ( 2 )

From ( 1 ) and ( 2 ),

$\Rightarrow$ $mx\pm \sqrt{m^2-3}=mx+\frac{2}{m}$

$\Rightarrow$ $\pm \frac{2}{3}=\sqrt{m^2-3}$

$\Rightarrow$ $\frac{4}{m^2}=m^2-3$

$\Rightarrow$ $4=m^4-3m^2$

Let $m^2=t$

$m^4=t^2$

$\Rightarrow$ $t^2-3t-4=0$

$\Rightarrow$ $t^2-4t+t-4=0$

$\Rightarrow$ $(t-4)(t+1)=0$

$\Rightarrow$ $t=4,-1$

But $-1$ is not possible.

$\Rightarrow$ $m^2=4$

$\therefore$ $m=\pm 2$

Equation of tangent when $m=2$

$\Rightarrow$ $y=mx+\frac{2}{m}$

$\Rightarrow$ $y=2x+1$

$\Rightarrow$ $2x-y+1=0$

When $m=-2$

$\Rightarrow$ $y=-2x-1$

$\Rightarrow$ $2x+y+1=0$

$\therefore$ Required equation is $2x\pm y+1=0$