Analytic geometry and calculus combined question

Question

Show that the equation for the tangent with the slope $m$, $(m≠0)$ to the parabola $y^2 = 4px$ is $y = mx + \frac{p}{m}$. How this is done? What is the method for proves of this kind?

Answer 1

Here is a simple solution that doesn't use calculus: We have $y^2 = 4px$ and $y = mx + \frac{p}{m}$. Substituting the second equation into the first, we find $$\left( mx + \frac{p}{m} \right)^2 = m^2x^2 + 2px + \frac{p^2}{m^2} = 4px \\ \Rightarrow m^2x^2 - 2px + \frac{p^2}{m^2} = 0 \\ \Rightarrow \left(mx - \frac{p}{m}\right)^2 = 0$$

Thus, the simultaneous equations have a single solution (with multiplicity $2$), which means that the line intersects the parabola at exactly one point, viz. $(x, y)$. Hence, the line in question must be the tangent, since any other line would pass through the parabola at two points or none, or at a single point different from $(x, y)$.

Answer 2

Hint

If you are allowed to use total differentiation, define the function as $$F(x,y)=y^2-4px=0$$ $p$ being a constant, the total differential is then $$dF(x,y)=2 y dy -4 p dx=0$$ from which you can extract $$\frac{{dy}}{{dx}}=\frac{2 p}{y}$$ which is the slope of the tangent.

I am sure that you can take from here.