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) = √( (x-2)^2 + 0)
square both sides and expand
x^2 + 8x + 16 + y^2 = x^2 - 4x + 4
y^2 = -12x - 12
Find the equation and sketch the parabola that has vertex V (-1; 0) and focus F(-4; 0).
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