Question

Find the equation of the common tangents to the parabola $y^2=8x$ and the hyperbola $3x^2-y^2=3$.

Answer 1

Any tangent $y^2=8x$ is $y=mx+\frac{2}{m}$ it should also be tangent to $3x^2-y^2=3$

$3x^2-{(mx+\frac{2}{m})}^2=3$ by substituting for $y=mx+\frac{2}{m}$

$\Rightarrow 3x^2-(m^2{x}^2+\frac{4}{m^2}+2\times mx\times \frac{2}{m})=3$

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

$\Rightarrow (3-m^2)x^2-4x-(\frac{4}{m^2}+3)=0$

Discriminant$=0\Rightarrow b^2-4ac=0$

$4^2-4\times (3-m^2)\times (\frac{4}{m^2}+3)=0$

$\Rightarrow 4[4+(3-m^2)\times (\frac{4}{m^2}+3)]=0$

$\Rightarrow 4+(3-m^2)\times (\frac{4}{m^2}+3)=0$

$\Rightarrow 4+9+\frac{12}{m^2}-3m^2-4=0$

$\Rightarrow -3m^2+\frac{12}{m^2}+9=0$

$\Rightarrow m^2-\frac{4}{m^2}-3=0$ by dividing the complete equation by $-3$

$\Rightarrow m^4-3m^2-4=0$ by multiplying the complete equation by $m^2$

$\Rightarrow (m^2-4)(m^2+1)=0$ by factorising

$\Rightarrow m^2-4=0$ and $m^2+1=0$

$\Rightarrow m^2=4\Rightarrow m=\pm 2$

$m^2=-1$ is not valid

Equation of the common tangents are $y=2x+\frac{2}{2}=2x+1$ and $y=-2x+\frac{2}{-2}=-2x-1$

$\therefore$ equation of common tangents are $y=2x+1$ and $y=-2x-1$