<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] 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 $m^2$
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