### Instruction: The parabola $y = x^2+2$ and the hyperbola $y^2 - mx^2 = 1$ are tangent. Find $m.$.### Response: I need to find a value of $m$ such that the parabola and the hyperbola have exactly one point of intersection.
To find the points of intersection, I can equate the $y$-values of the two curves and solve for $x$.
So I have $x^2+2 = \sqrt{1+mx^2}$, which I can square both sides and rearrange to get $x^4 + (4-m)x^2 + 3 = 0$.
This is a quartic equation, but I can treat it as a quadratic equation in $x^2$ and use the quadratic formula.
Let $u = x^2$. Then I have $u^2 + (4-m)u + 3 = 0$.
The discriminant of this equation is $(4-m)^2 - 12$, and I want this to be zero, because that means there is only one solution for $u$.
So I set $(4-m)^2 - 12 = 0$ and solve for $m$.
I get $m = 4 \pm 2\sqrt{3}$.
However, I need to check if these values of $m$ actually give me a valid point of intersection.
I can plug them back into the equation for $u$ and get $u = \frac{m-4}{2}$.
If $m = 4 + 2\sqrt{3}$, then $u = \sqrt{3}$, and $x = \pm \sqrt{\sqrt{3}}$.
If $m = 4 - 2\sqrt{3}$, then $u = -\sqrt{3}$, and $x$ is imaginary.
[* ] I only want real values of $x$, so I reject the second value of $m$.
Therefore, the only value of $m$ that works is $m = 4 + 2\sqrt{3}$.