Prove that the lines $\displaystyle y=mx\pm a\sqrt{m^2-1}$ are tangent to the hyperbola $\displaystyle x^2-y^2=a^2$ for all values ofm.

My work:

I substituted $\displaystyle y=mx\pm a\sqrt{m^2-1}$ into the equation.

$\displaystyle x^2-m^2x^2\mp 2am\sqrt{m^2-1}x-a^2(m^2-1)=a^2$

$\displaystyle (1-m^2)x^2\mp 2am\sqrt{m^2-1}x-a^2(m^2-2)=0$

If $\displaystyle y=mx\pm a\sqrt{m^2-1}$ is a tangent to the hyperbola,

$\displaystyle b^2=4ac$

$\displaystyle \rightarrow(2am\sqrt{m^2-1})^2=-4(1-m^2)(m^2-2)$

I solved formand got $\displaystyle \pm 1$

So this is definitely wrong, but I don't know what is wrong.

Thanks for any help!