Write an equation for each parabola

Question

(a)

vertex (2,-1)

focus (2,-5)

(b)

focus (3,5)

directrix x=0

Answer 1

Ignoring offsets, remember that the basic equation is one of 4 possibilities:

y² = 4ax opens RIGHT

y² = -4ax opens LEFT

x² = 4ay opens UP

x² = -4ay opens DOWN

a is the distance from the vertex to either the focus or directrix

(a)

vertex (2,-1)

focus (2,-5)

The x values are the same, so it opens UP or DOWN. The focus is more negative than the vertex, so it opens down. a = 5-1 = 4, giving us the basic equation x² = -4ax = -16x

Finally, put in the offsets: since the vertex is at (2,-1), the offsets are -2 and +1......SO:

(x - 2)² = -16(y + 1)

Answer 2

(a)

vertex (2,-1)

focus (2,-5)

This parabola must be concave down since the two points are on the same vertical line x=2 and the vertex is above the focus.

So we use the general form (x-h)²=4p(y-k). We are given the vertex is (2,-1)=(h,k) so

(x-2)²=4p(y--1)

(x-2)²=4p(y+1)

And we are given the focus (h,k+p)=(2,-5) hence k+p=-5 and p=-5-k=-5--1=-4

Finally, (x-2)²=4p(y+1)

(x-2)²=4(-4)(y+1)

(x-2)²=-16(y+1)

(b)

focus (3,5)

directrix x=0

The directrix is always perpendicular to the vertex's axis of symmetry. So since x=0 is the directrix, the vertex's axis of symmetry must be sideways, meaning the parabola is concave-right or concave-left. In this case, since the focus (inside the "cup") is at (3,5), the parabola is concave right (3>0).

So we start with the general form (y-k)²=4p(x-h) for a concave-horizontal parabola.

(h+p,k) = (3,5) so k = 5 and h+p = 3.

We can now write:

(y-5)² = 4p(x-h)

the directrix x = h-p = 0 so h = p.

Since we know h+p = 3, p+p = 3, 2p = 3, p=3/2=h

(y-5)² = 4(3/2)(x-3/2)

(y-5)² = 6(x-3/2)