What is the vertex form of the equation of the parabola with a focus at (55,45) and a directrix of #y=47 #?

1 Answer
Mar 31, 2018

#y=-1/4(x-55)^2+46#

Explanation:

Parabola is locus of a point which moves so that its distance from a given point calld focus and a given line ccalled directrix is always constant.

Let the point be #(x,y)#. Here focus is #(55,45)# and distance from focus is #sqrt((x-55)^2+(y-45)^2)#. Its distance from directrix #y=47# i.e. #y-47=0# is #|y-47|#.

Hence equaion of parabola is

#(x-55)^2+(y-45)^2)=|y-47|^2#

or #(x-55)^2+y^2-90y+2025=y^2-94y+2209#

or #(x-55)^2-184=-4y#

i.e. #y=-1/4(x-55)^2+46#

Hence parabola's equation is #y=-1/4(x-55)^2+46# and vertex is #(55,46)#.

graph{((x-55)^2-184+4y)(y-47)((x-55)^2+(y-45)^2-0.1)=0 [35, 74.99, 35.02, 55.02]}