<s>[INST] A supposed to be easy calculus problem Find the values of $m$ if the line $y=mx+2$ is a tangent to the curve $x^2-2y^2=1$.
My working:
First we differentiate $x^2-2y^2=1$ with respect to $y$ to get the gradient. We get $y^2=\frac{1}{2}x^2-\frac{1}{2}\implies y=\pm\sqrt{\frac{1}{2}x^2-\frac{1}{2}}$.
We take the positive one for demonstration
$\frac{dy}{dx}=\frac{1}{2}x(\frac{1}{2}x^2-\frac{1}{2})^{-\frac{1}{2}}=\frac{x}{2\sqrt{\frac{1}{2}x^2-\frac{1}{2}}}$
$\implies(1-2m^2)x^2=-2m^2$
Since the tangent touches the curve, we can make $x^2-2(mx+2)^2=1$, we then get $(1-2m^2)x^2=9+8mx$
$\implies(1-2m^2)x^2=-2m^2$ and $(1-2m^2)x^2=9+8mx$ are two equations with two unknowns, then we should be able to find the values of $m$, but I couldn't find any easy way to solve those 2 simultaneous equations. Is there any easier method?
I tried solving $9+8mx=-2m^2$ but we still have two unknowns in one equation?
Also, if we don't use those two simultaneous equations, can we solve this question with a different method?
I am trying to solve WITHOUT implicit differentiation.
Many thanks for the help!
 [/INST] Let the given line be tangent to the curve at the point $(x_0, y_0).$ Then we have $$y_0 = m x_0 + 2, \\ x_0^2 = 1 + 2y_0^2.$$ Using the fist equation in the second, we have $$x_0^2 = 1 + 2 (m x_0^2 + 4 m x_0 + 4),$$ which is $$m = \frac {x_0^2 - 9} {2 x_0^2 + 8 x_0}.$$ We also know from the second equation that $$y' = \frac{x} {\sqrt{2x^2-2}}.$$ Hence, $$m = \frac{x_0} {\sqrt{2x_0^2-2}}.$$ Solving the equation $$\frac{x_0} {\sqrt{2x_0^2-2}} = \frac {x_0^2 - 9} {2 x_0^2 + 8 x_0},$$ we get $x_0 = -3.44122\cdots$ and $m = -.73899\cdots$. 
</s>