Finding an equation of the parabola and the value of x with the equation given.

Question

A parabola has a focus of (0,3) and a vertex at the origin. Write an equation of the parabola. Then find the value of x for which y-8=0.

Answer 1

The general equation for a parabola is

(x - h)²=4p(y - k)

where (h,k) is the vertex of the parabola. In the case of your problem, the vertex is (0,0), so you don't have to take those factors into account and the equation can be simplified to

x²=4py

p is found by finding the distance between the vertex and the focus, or 3 - 0 = 3.

x²=12y or y= x²/12

---

for y-8=0, the equation of the line is y=8. The y value is 8 for all values of x, and this is a horizontal line at y=8. This line would cross the parabola whenever y =8. For a parabola, this will yield two values.

Solve x²=12y for y=8 to get that x=±9.8

Answer 2

vertex = (0,0)

focus = (0,3)

when the focus is directly above the vertex, with the same x coordinate, the parabola opens upward with the

general form y = ax^2

the directrix line is the same distance below the vertex as the focus is above the vertex

or y=-3

the parabola is the collection of all points (x,y) such that the distance is the same from

the parabola to the focus as to the directrix

pick a point on the parabola horizontal from the focus (x,3)

the distance from (x,3) to the focus (0,3) = x = the distance from (x,3) to the directrix y=-3

from (x,3) to y= -3 is 6, so x = 6

Now you have a point on the parabola other than the vertex, (6,3)

plug that point into y=ax^2 to find the value of a

3 =a(6)^2 = 36a

36a = 3

a = 3/36 = 1/12

the parabola equation is

y= (1/12)x^2

since a=1/12

to remove the fraction, multiply both sides by 12

to get

12y=x^2

the standard parabola y=x^2 has a=1

as a gets very large, the parabola gets very narrow, vertical, getting closer to two vertical lines

as a gets closer to zero, the parabola gets flatter, closer to a horizontal line

1/12 is close to zero, so the parabola is fairly flat

find a value of x for when y-8 = 0?

If y-8 = 0, then y=8

y=(1/12)x^2

8 = (1/12)x^2

x^2 = 12(8) = 96

x =sqr96 = 4sqr6 = 4 times the square root of 6 = about 9.8

y=(1/12)sqr96 squared = (1/12)96 = 96/12 = 8