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#