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

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

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

-(1/40)x^2=y

or as or still equivalently .

Compare that green outlined equation to the basic form, .

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