Question

How do you show that the point `(p,4p^2)` lies on the curve `y=4x^2` for all real values of p and then find the equation of the tangent to `y=4x^2` at `(p,4p^2)`?

Answer 1

You can substitute the coordinate `x=p` of your point into the function to get: `y=4p^2` so your point does lie on the curve represented by `y=4x^2`.

Your function is a quadratic which can accept all real values of `x` so that `p` can be every real value.

Deriving your function you get the slope of the tangent to the curve at a generic `x`:
`y'=8x`

The slope in `x=p` is: `y'(p)=8p`

The equation of the line through your point (of coordinates `(x_0,y_0)`) and slope `m=8p` is:
`y-y_0=m(x-x_0)`
`y-4p^2=8p(x-p)`
and finally:
`y=8px-4p^2`