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!
 A: One more simplest way:
Put $y=mx+2$ in the equation $x^2-2y^2=1$. Then it comes to a quadratic equation of $x$. From which we get two values of $x$. Since the line is tangent to the given hyperbola so, it can not intersect at two different points. So, the quadratic equation must give two identical values of $x$.
For this,  put discriminant is equal to $0$. 
Quadratic equation becomes , $x^2-2(mx+2)^2=1$. Putting discriminant equal to $0$ we get, $$64m^2+36(1-2m^2)=0\implies m=\pm \frac{3}{\sqrt 2}$$
A: Let $(a,b)$ be a point of tangency. We have $2x-4y\frac{dy}{dx}=0$, so the slope of the tangent line at $(a,b)$ (if $b\ne 0$) is $\frac{a}{2b}$.
The tangent line has equation $y-b=(x-a)(a/2b)$. Simplifying , and comparing with $y=mx+2$, we find that $b-a^2/(2b)=2$. It follows that $2b^2-a^2=4b$. Since $a^2-2b^2=1$, we conclude that $b=-1/4$. The rest is routine.
Remark: Note that the tangent line happens to be at a point on the "lower" half of the hyperbola. So taking the positive square root turns out not to be useful. 
A: You have $$(1-2m^2)x^2=-2m^2\\(1-2m^2)x^2=9+8mx\\-2m^2=9+8mx\\x=\frac{-2m^2-9}{8m}\\x^2=\frac{-2m^2}{1-2m^2}\\\frac{4m^4+36m^2+81}{64m^2}=\frac{-2m^2}{1-2m^2}$$Now you can use the quadratic equation in $m^2$
A: 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 $m = -.73899\cdots$. 
