What is the equation of a parabola with a focus at (-2, 6) and a vertex at (-2, 9)? What if the focus and vertex are switched?

1 Answer
Jul 21, 2018

The equation is #y=-1/12(x+2)^2+9#. The other equation is #y=1/12(x+2)*2+6#

Explanation:

The focus is #F=(-2,6)# and the vertex is #V=(-2,9)#

Therefore, the directrix is #y=12# as the vertex is the midpoint from the focus and the directrix

#(y+6)/2=9#

#=>#, #y+6=18#

#=>#, #y=12#

Any point #(x,y)# on the parabola is equidistant from the focus and the directrix

#y-12=sqrt((x+2)^2+(y-6)^2)#

#(y-12)^2=(x+2)^2+(y-6)^2#

#y^2-24y+144=(x+2)^2+y^2-12y+36#

#12y=-(x+2)^2+108#

#y=-1/12(x+2)^2+9#

graph{(y+1/12(x+2)^2-9)(y-12)=0 [-32.47, 32.45, -16.23, 16.25]}

The second case is

The focus is #F=(-2,9)# and the vertex is #V=(-2,6)#

Therefore, the directrix is #y=3# as the vertex is the midpoint from the focus and the directrix

#(y+9)/2=6#

#=>#, #y+9=12#

#=>#, #y=3#

#y-3=sqrt((x+2)^2+(y-9)^2)#

#(y-3)^2=(x+2)^2+(y-9)^2#

#y^2-6y+9=(x+2)^2+y^2-18y+81#

#12y=(x+2)^2+72#

#y=1/12(x+2)^2+6#

graph{(y-1/12(x+2)^2-6)(y-3)=0 [-32.47, 32.45, -16.23, 16.25]}