Proving that a quartic has exactly 2 real roots **Question:** Find the equation of the normal to the hyperbola $xy=c^2$ at $P(ct,\frac{c}{t})$ then prove that _exactly_ two normals can be drawn to the hyperbola $xy=c^2$ from a point $(0,k)$, where $k$ is real. **Attempt** I fount the equation of the normal. It is $ty-t^3x=c(1-t^4)$ Sub $x=0$ and $y=k$ to get: $ct^4+kt-c=0$ Now I need to prove that this equation has _exactly_ 2 real roots. How can I do that?

Hadn't someone already mentioned Descartes' Rule of Signs? Maybe I imagined it. Anyway ...

Take $k$ and $c$ non-zero (the zero cases are trivial), and name the polynomial $p(t)$.

* The coefficient sequence of $p(t)$ is $(c, k, −c)$, which has one sign change ($k$ matches sign with either $c$ or $−c$). Thus, by the Rule of Signs, $p$ has exactly one positive real root.
* The coefficient sequence for $p(−t)$ is $(c, −k, −c)$, which also has one sign change, so that $p$ has exactly one negative root.



Since zero isn't a root, and since we've run out of real candidates, the remaining roots must be non-real.