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

Oct 2, 2017

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

Explanation:

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

$d = x - 1 \text{ [1]}$

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

$d = \sqrt{{\left(x - \left(- 1\right)\right)}^{2} + {\left(y - 0\right)}^{2}}$

Simplify:

$d = \sqrt{{\left(x + 1\right)}^{2} + {y}^{2}} \text{ [2]}$

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

$x - 1 = \sqrt{{\left(x + 1\right)}^{2} + {y}^{2}}$

Square both sides:

${\left(x - 1\right)}^{2} = {\left(x + 1\right)}^{2} + {y}^{2}$

Expand the squares:

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

Combine like terms:

$- 4 x = {y}^{2}$

Divide both sides by -4:

$x = - \frac{1}{4} {y}^{2} \leftarrow$ standard form for a parabola that opens left.