Question

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

Answer 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

Answer 2

`y=(-1)/(4)x^2`

Explanation:

The general equation of parabola :

`(x -h)^2= 4p(y -k)` (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) :
`x^2= -4y`

`i.e.\ y=(-1)/(4)x^2`