What is the equation of the parabola with a focus at (3,-8) and a directrix of y= -5?

2 Answers
Jun 26, 2018

The equation is #y=-1/6(x-3)^2-39/6#

Explanation:

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

Therefore,

#(y+5)=sqrt((x-3)^2+(y+8)^2)#

Squaring both sides

#(y+5)^2=(x-3)^2+(y+8)^2#

#y^2+10y+25=(x-3)^2+y^2+16y+64#

#6y=-(x-3)^2-39#

#y=-1/6(x-3)^2-39/6#

graph{(y+1/6(x-3)^2+39/6)(y+5)=0 [-28.86, 28.87, -14.43, 14.45]}

Jun 26, 2018

The equation of parabola is #y=-1/6(x-3)^2-6.5 #

Explanation:

Focus is at #(3,-8) #and directrix is #y=-5#. Vertex is at midway

between focus and directrix. Therefore,vertex is at #(3,(-5-8)/2)#

or at #(3, -6.5)# . The vertex form of equation of parabola is

#y=a(x-h)^2+k ; (h,k) # being vertex. # h=3 and k = -6.5#

So the equation of parabola is #y=a(x-3)^2-6.5 #. Distance of

vertex from directrix is #d= |6.5-5|=1.5#, we know # d = 1/(4|a|)#

#:. 1.5 = 1/(4|a|) or |a|= 1/(1.5*4)=1/6#. Here the directrix is above

the vertex , so parabola opens downward and #a# is negative.

#:. a= -1/6#. Hence the equation of parabola is

#y=-1/6(x-3)^2-6.5 #

graph{-1/6(x-3)^2-6.5 [-40, 40, -20, 20]}