# How do you find an equation of the parabola with vertex (-2,1) and directrix x=1?

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6
Sep 22, 2017

$x = - {y}^{2} / 6 + \frac{y}{3} - \frac{13}{6}$

#### Explanation:

Given -

Vertex $\left(- 2 , 1\right)$
Directrix $x = 1$

The vertex is in the 2nd quadrant. The directrix is parallel to the y-axis. So, the parabola opens to the left. The vertex of the parabola is not the origin. Then its general form is -

${\left(y - k\right)}^{2} = - 4. a . \left(x - h\right)$

Where -
$h$ and $k$ are the coordinates of the vertex.

h=-2)

$k = 1$

$a = 1.5$ half the distance between Directrix and vertex [= distance between focus and vertex]

Substitute these values in the equation

${\left(y - 1\right)}^{2} = - 4.1 .5 . \left(x + 2\right)$
${y}^{2} - 2 y + 1 = - 6 x - 12$
$- 6 x - 12 = {y}^{2} - 2 y + 1$
$- 6 x = {y}^{2} - 2 y + 1 + 12$
$x = {y}^{2} / \left(- 6\right) - \frac{2 y}{- 6} + \frac{13}{- 6}$

$x = - {y}^{2} / 6 + \frac{y}{3} - \frac{13}{6}$

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