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

##### 1 Answer

Sep 22, 2017

#x=-y^2/6+y/3-13/6#

#### Explanation:

Given -

Vertex

Directrix

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 -

#(y-k)^2=-4.a.(x-h)#

Where -

#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

#(y-1)^2=-4.1.5.(x+2)#

#y^2-2y+1=-6x-12#

#-6x-12=y^2-2y+1#

#-6x=y^2-2y+1+12#

#x=y^2/(-6)-(2y)/(-6)+13/(-6)#

#x=-y^2/6+y/3-13/6#