1. Tangent to hyperbola

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

My work:
I substituted $y=mx\pm a\sqrt{m^2-1}$ into the equation.
$x^2-m^2x^2\mp 2am\sqrt{m^2-1}x-a^2(m^2-1)=a^2$
$(1-m^2)x^2\mp 2am\sqrt{m^2-1}x-a^2(m^2-2)=0$
If $y=mx\pm a\sqrt{m^2-1}$ is a tangent to the hyperbola,
$b^2=4ac$
$\rightarrow(2am\sqrt{m^2-1})^2=-4(1-m^2)(m^2-2)$
I solved for m and got $\pm 1$
So this is definitely wrong, but I don't know what is wrong.
Thanks for any help!

2. Check your b^2 = 4ac calculation.

3. this is my working:
$4a^2m^2(m^2-1)=-4a^2(1-m^2)(m^2-2)$
$4a^2m^4-4a^2m^2=-4a^2(3m^2-2-m^4)$
$4a^2m^4-4a^2m^2=-12a^2m^2+8a^2+4a^2m^4$
$8a^2m^2=8a^2$
$m^2=\frac{8a^2}{8a^2}=1$
$m=\pm 1$

4. The last term is -a^2*m^2 + a^2 not -a^2*m^2 -a^2

5. huh? i went through all my workings again and could not see any mistakes which when corrected would give me that result.

6. Originally Posted by arze
huh? i went through all my workings again and could not see any mistakes which when corrected would give me that result.

The above equation becomes
x^2(1-m^2) + 2amsqrt(m^2-1) -a^2*m^2 = 0
In this b^2 = 4ac.

7. oh! ok i see where i got it wrong. thanks!