SOLUTION: Find the vertex, focus, directrix, and focal width of the parabola. -1/40 x^2= y

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the vertex, focus, directrix, and focal width of the parabola. -1/40 x^2= y      Log On


   

Question 983230: Find the vertex, focus, directrix, and focal width of the parabola.
-1/40 x^2= y
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
-1/40 x^2= y

-(1/40)x^2=y

-%281%2F40%29x%5E2=y or as x%5E2=-40y or still equivalently highlight_green%28-40y=x%5E2%29.

Compare that green outlined equation to the basic form, 4p=x%5E2.

A video which explains this well (but a little bit clumsy) is this one:
Equation Derivation for Parabola from Focus and Directrix