Write an equation in standard form for a parabola with focus (−3, −10), that opens down, and contains the point (5, −10).

Question

I need help determining h, k, and p.

Answer 1

downward opening parabolas are of the form

y =-a(x+h)^2 + k where (h,k) is the vertex, if (-3,-10) is the focus, axis of symmetry is x=-3 and h=-3

y =-a(x+3)^2 + k where k>-10 and a>0

(5,-10) is a point on the parabola and 8 units from the focus. It's also 8 units to the directrix or y=-2

the vertex is half way between the focus and directrix, so it's (-3,-6)

y=-a(x+3)^2 -6. plug in the point (5,-10) to solve for a

-10 =-a(8^2)-6

a = -4/64 = 1/16

the parabola is y=(-1/16)(x+3)^2 - 6

the coefficient of the x^2 term is 1/4p p is the distance from the vertex to focus = -4

1/4p = -1/16

h=-3, k= -6, p= -4