How do you find the standard form of the equation Vertex: ( -1, 2), Focus (-1, 0) Vertex: (-2, 1), Directrix: x = 1?

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Answer 1 · 2017-10-02T18:29:42

Use the fact that a parabola is the locus of points equidistant from the focus point and the directrix line.

The distance from the directrix, $x =1$ , to any point, $(x,y)$ , on the parabola is:

$d = x - 1" [1]"$

The distance from the focus, $(-1,0)$ to any point, $(x,y)$ , on the parabola is:

$d = sqrt((x-(-1))^2+(y-0)^2)$

Simplify:

$d = sqrt((x+1)^2+y^2)" [2]"$

Because the distances must be equal, we can set the right side of equation [1] equal to the right side of equation [2]:

$x -1 = sqrt((x+1)^2+y^2)$

Square both sides:

$(x -1)^2 = (x+1)^2+y^2$

Expand the squares:

$x^2-2x+1 = x^2+2x+1+y^2$

Combine like terms:

$-4x = y^2$

Divide both sides by -4:

$x = -1/4y^2 larr$ standard form for a parabola that opens left.