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?
 A: 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.
A: Take $$f(t)=ct^4+kt-c$$
At t=0 we have $$f(t)=-c$$
Also, $$\displaystyle\lim_{t\to \pm \infty} = \infty $$ so by intermediate value theorem there exist at least 2 real roots; one positive, one negative.  
Edit: 
Now consider the derivative: $$f'(t) = 4ct^3+k$$
This is clearly strictly increasing, so there is only one turning point of f(t), hence exactly two roots.
A: I think I got it.
Let the roots of $ct^4+kt-c$ be $\alpha, \beta, \gamma, \delta$
$\sum \alpha =-\frac{0}{c}=0$
$\sum \alpha \beta =\frac{0}{c}=0$
$\sum \alpha ^{2}=\left ( \sum\alpha \right )^{2}-2\sum \alpha \beta =0-2(0)=0$
Therefore, there must be at least one imaginary root in $\alpha ^{2}+\beta^{2}+\gamma^{2}+\delta^{2}$ because it is impossible to get zero if all the roots are real. Since the coefficients are real, by the complex conjugate root theorem, there must be two imaginary roots.
To prove that not all the roots are imaginary, we can use FireGarden's answer.
Is this correct?
