parabolas
posted by dylan .
Find the equation and sketch the parabola that has vertex V (1; 0) and focus F(4; 0).

The distance between the vertex and the focus is 3
so the directrix must be x =2 (from 1 to 2 is 3)
let (x,y) be any point on our parabola
√( (x+4)^2 + y^2) = √( (x2)^2 + 0)
square both sides and expand
x^2 + 8x + 16 + y^2 = x^2  4x + 4
y^2 = 12x  12
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