Question

The equation(s) of the common tangent(s) to the parabola $y^2=8x$ and $3x^2-y^2=3$ is /are :

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

Answer 1

Answer: (A), (C)

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

The equation of tangent to parabola $y^2=8x$ in slope form is given by
$y=mx+\frac{2}{m}$ ... (i).
This is also tangent to the given hyperbola $3x^2-y^2=3$.
$\Rightarrow 3x^2-(mx+\frac{2}{m}{)}^2=3$
$x^2(3-m^2)-4x-\frac{4}{m^2}-3=0$
For roots to be equal, discriminant of the above quadratic equation must be zero, then
$16-(-\frac{4}{m^2}-3)(3-m^2)=0$
On solving, we get
$4m^4-3m^2-4=0$
$(m^2+1)(m^2-4)=0$
$m^2=-1$ and $m^2=4$
Since $m^2$ can not be negative, therefore $m^2=4$
$m=-2,2$
Substituting the value of $m$ in equation $\left(i\right)$, we get
$y=2x+1$ and $y=-2x-1$
or $2x-y+1=0$ and $2x+y+1=0$
Hence, options 'A' and 'C' are correct.