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

Answer:  

$y=\dfrac{-1}{4}x^2$

Step-by-step 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) : $x^2= -4y$

$i.e.\ y=\dfrac{-1}{4}x^2$