Write an equation of a parabola that opens downward, has a vertex at the origin, and a focus at (0, –1).

y=a(x-h)² + k, vertex(0,0)
(h, k) vertex, h=0, k=0
y=a(x-0)²+0=ax
y=ax²
focus(h, k+ 1/4*a) focus (0, -1)
k+ 1/4 * a= - 1,
0+1/4 a=-1
1/4a=-1
a=-4
y =ax². y=-4x²

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JeanaShupp JeanaShupp · Answer 2 · 2020-03-24T03:34:40+01:00

Answer:

[tex]y=\dfrac{-1}{4}x^2[/tex]

Step-by-step explanation:

The general equation of parabola :

[tex](x -h)^2= 4p(y -k)[/tex] (i)

where (h,k) = vertex of parabola

Focus = (h,k+p)

When p<0 , the parabola opens downwards.

As per given , (h,k) = (0,0)

focus = (0, –1).

i.e. h=0 , k=0 , k+p=-1 , i.e. p=-1

On substituting the value of h , k and p in (i) , we get

Equation of a parabola that opens downward, has a vertex at the origin, and a focus at (0, –1) : [tex]x^2= -4y[/tex]

[tex]i.e.\ y=\dfrac{-1}{4}x^2[/tex]