<s>[INST] Find the point where L intersects the curve The curve $x^2 - 2y^2 = 1$ includes the point $(1, 0)$. Let $L$  be the line through $(1, 0)$ having slope $m$. Find the other point where $L$ intersects the curve.
The line can be written as $y=m(x-1)$. Substituting this value of y in the original curve gives us:
$x^2(1-2m^2) + 4m^2x + 2m^2 - 1 = 0$.
I think we can divide this equation by $x-1$ so we get another solution but I couldn't do that. Can someone help me?
 [/INST] Note the equation should actually be $$x^2(1-2m^2) +4m^2x \color{red} - 2m^2 -1=0$$  Performing long division by $(x-1)$ yields $$(x-1) \left( (1-2m^2)x +(2m^2+1) \right) =0 \\ \implies x=1, \frac{2m^2+1}{2m^2-1}$$  and the required point is $$\left ( \frac{2m^2+1}{2m^2-1} , \frac{2m}{2m^2-1} \right)$$
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