Quadratics - the Australian Mathematical Sciences Institute

A guide for teachers – Years 11 and 12 • {23}
y
S(0,a)
Focus
0
Directrix
x
y = –a
We now look at the locus of a point moving so that its distance from the focus is equal to
its perpendicular distance to the directrix. Let P(x, y) be a point on the locus. Our aim is
to find the equation governing the coordinates x and y of P.
y
P(x,y)
S(0,a)
Focus
PT = PS
0
Directrix
T
a
x
y = –a
The distance PS is equal to √ x 2 + (y − a) 2 , and we can see from the diagram that the
perpendicular distance PT of P to the directrix is simply y + a.
Since PS = PT , we can square each length and equate, giving
x 2 + (y − a) 2 = (y + a) 2
=⇒ x 2 = 4ay.
The last equation may also be written as
y = x2
4a
and so we see that it is the equation of a parabola with vertex at the origin.
The positive number a is called the focal length of the parabola.
Any parabola of the form y = Ax 2 + Bx +C can be put into the standard form
(x − p) 2 = ±4a(y − q),
with a > 0, where (p, q) is the vertex and a is the focal length. When a parabola is in this
standard form, we can easily read off its vertex, focus and directrix.