What is the vertex form of the equation of the parabola with a focus at (-4,-7) and a directrix of #y=10#?

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Answer 1 · 2018-05-07T10:53:33

The equation of parabola is #y=-1/34(x+4)^2+1.5 #

Focus is at #(-4,-7) #and directrix is #y=10#. Vertex is at midway

between focus and directrix. Therefore vertex is at

#(-4,(10-7)/2)or (-4, 1.5)# . The vertex form of equation of

parabola is #y=a(x-h)^2+k ; (h.k) ;# being vertex.

# h= -4 and k = 1.5#. So the equation of parabola is

#y=a(x+4)^2 +1.5 #. Distance of vertex from directrix is

#d= 10-1.5= 8.5#, we know # d = 1/(4|a|)#

#:. 8.5 = 1/(4|a|) or |a|= 1/(8.5*4)=1/34#. Here the directrix is

above the vertex , so parabola opens downward and #a# is

negative #:. a=-1/34# Hence the equation of parabola is

#y=-1/34(x+4)^2+1.5 #

graph{-1/34(x+4)^2+1.5 [-40, 40, -20, 20]}