Question

How do I find the equation of a parabola given the vertex and a point on the parabola?

One of these problems gives me vertex (0,1) and point (2,2).

The book says the answer is x^2 = 4(y-1)

(I really don't know where to start because everything I've tried so far hasn't worked.)

Answer 1

the general form of a parabola is

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

where

(h,k) is the vertex
p is the focal distance (ie the distance from the vertex to the focus along the axis of symmetry)

Answer 2

the vertex is given to be (0,1)

so

(h,k) = (0,1)

which means

h = 0
k = 1

Answer 3

plug in h = 0 and k = 1 to get

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

(x-0)^2 = 4p(y-1)

x^2 = 4p(y-1)

Answer 4

then plug in the other point (x,y) = (2,2), and solve for p, to get


x^2 = 4p(y-1)

2^2 = 4p(2-1)

4 = 4p(1)

4 = 4p

4/4 = p

1 = p

p = 1

Answer 5

So,

x^2 = 4p(y-1)

turns into

x^2 = 4*1(y-1)

x^2 = 4(y-1)

Answer 6

Thank you!

Answer 7

you're welcome

Answer 8

Jim used the useful form of the equation, but sometimes you see "vertex form" written as

\[ y = a(x-h)^2 + k \]
(which is the same a Jim's but does not show the "p" )
h and k are the vertex, so you would replace (h,k) with (0,1)
\[ y = a(x-0)^2 + 1 \\ y = ax^2 + 1\]
you would use the known point (2,2) to find a:
\[ 2= a\cdot 2^2 + 1 \\ 1 =4a \\ a= \frac{1}{4} \]
and
\[ y = \frac{1}{4}x^2 + 1 \]
which is "vertex form" of the equation

Answer 9

Ah, I see. Thank you!