Given a parabola $y^2=2px$ we must prove that the tangent to the parabola at $P_1(x_1,y_1)$ is of the form $yy_1=p(x+x_1)$.
One way to do this is using derivatives. Another way is using Archimedes' Lemma. A third way is the one I tried to do. I've looked up a lot for this particular proof but I wasn't able to find exactly this anywhere I've searched.
I tried to prove this using a system of equations. Tangent: $y=mx+β$, where $β=-mx_1+y_1$ and parabola: $y^2=2px$. Substituting the equation of the tangent in the equation of the parabola and setting $Δ=0$ for one root, after some algebra we get: $-2x_1m^2+2y_1m-p=0$
How is it possible that this is a quadratic equation if we only have one tangent line? In a similar problem with a hyperbola and a tangent we get a quadratic equation as there are two tangents, one for each part of the curve.
Which slope $m$ should I choose and why?
Thanks in advance!