Question

Parabola question!

Answer 1

How do you use the general equation of a tangent \[y=tx-at^2\]

Answer 2

this is the equation of the tangent at the point (2at, at^2)


If the parabola is \[x^2 = 4ay \]


then \[y = \frac{x^2}{4a}...so... y' = \frac{x}{2a}\]

find the gradient at the point (2at, at^2) substitute x = 2at

so the gradient is

\[m = \frac{2at}{2a} ..... m = t\]

so using the point slope formula

\[y - at^2 = t(x - 2at)\]

which gives the equation of the tangent you have started.

not sure what you are really asking for. Perhaps this helps.