# Tangent to parabola without derivative

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?

• How do you express the equation of the tangent at the origin in the form $y=mx+\beta$? – amd Mar 24 '17 at 17:32
• @amd I don't know that, how? (the final equation for the tangent still works for the origin though..) – Michalis P. Mar 24 '17 at 17:40
• You can’t, which is my point. Your proof is incomplete because it doesn’t cover that case. – amd Mar 24 '17 at 17:41
• @amd So how do I prove that this is the equation in this case as well? – Michalis P. Mar 24 '17 at 17:44
• You’ll either have to handle it as a special case or start with a different form of equation of a line that can express vertical lines as well. – amd Mar 24 '17 at 17:45

Observe that, in the equation, $$-2x_1m^2+2y_1m-p=0$$ The discriminant $\Delta=(2y_1)^2-4\cdot(-2x_1)\cdot (-p)=4y_1^2-8px_1$.
And since $(x_1,y_1)$ lies on the parabola $y^2=2px$, so $y_1^2=2px_1 \implies 4y_1^2=8px_1$.
So the discriminant $\Delta=0$.
This implies only one real value of $m$.