What is the equation of the parabola with a focus at (-3,1) and a directrix of y= 0?

Question

Kennedy Adams · Accepted Answer

The equation of parabola is #y=1/2(x+3)^2+0.5 #

Focus is at #(-3,1) #and directrix is #y= 0#. Vertex is at midway

between focus and directrix. Therefore vertex is at #(-3,(1-0)/2)#

or at #(-3, 0.5)# . The vertex form of equation of parabola is

#y=a(x-h)^2+k ; (h.k) ;# being vertex. # h=-3 and k = 0.5#

Therefore vertex is at #(-3,0.5)# and the equation of parabola is

#y=a(x+3)^2+0.5 #. Distance of vertex from directrix is

#d= 0.5-0=0.5#, we know # d = 1/(4|a|) :. 0.5 = 1/(4|a|)# or

#|a|= 1/(4*0.5)=1/2#. Here the directrix is below

the vertex , so parabola opens upward and #a# is positive.

#:. a=1/2# . The equation of parabola is #y=1/2(x+3)^2+0.5 #

graph{1/2(x+3)^2+0.5 [-5, 5, 10, 10]} [Ans]

Jace Anderson · Answer

undefined The equation of the parabola with a focus at (-3,1) and a directrix of y= 0 is:

$ (x + 3)^2 + (y - 1)^2 = (y - 0)^2 $