Question

Find the standard form of the equation of the parabola with a focus at (0, 8) and a directrix at y = -8.

Answer 1

vertical

Answer 2

#1 - You MUST be able to find the Vertex from this information. Do tell!

Answer 3

I think it is (0,0)

Answer 4

Are you SURE? What is your evidence?

Answer 5

the answer options

Answer 6

Boo!!! Quite unsatisfactory. The DEFINITION of a parabola suggests that the Vertex is the midpoint of the line segment constructed between the Focus and perpendicular to the Directrix. What is the midpoint of the segment with endpoints (0,8) and (0,-8)?

Answer 7

ok so the distance from directrix to focus is the same distance from vertex to focus right?

Answer 8

(0,0)

Answer 9

That's it. No more guessing.

The VERTEX form of a parabola, with vertical orientation is

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

We know already that the Vertex is (0,0). This simplifies life quite a bit, leaving...

\(x^{2} = 4py\)

What is "p"? It also has something to do with those distances we were talking about.

Answer 10

16 is the distance from focus to directrix

Answer 11

Right. And half of that, is p.

Answer 12

x^2=8y

Answer 13

No, that's just "p". You need "4p".

Answer 14

oops 4*8

Answer 15

0,0

Answer 16

You can also prove it with the strict definition.

Distance of a point from the Focus: \(\sqrt{x^{2}+(y-8)^{2}}\)
Distance of a point from the Directrix: y + 8

These must be equal for all points on the parabola.

\(\sqrt{x^{2}+(y-8)^{2}} = y + 8\)

\(x^{2} + (y-8)^{2} = (y + 8)^{2}\)

\(x^{2} + y^{2} - 16x + 64 = y^{2} + 16y + 64\)

\(x^{2} = 32y\)

It seemed important to do it both ways and PROVE it, since this does not seem to be amongst your choices.

Answer 17

Oh, and this is NOT the STANDARD FORM. This is the Vertex Form.

STANDARD FORM would be perhaps \(x^{2} + 32y = 0\).

This problem statement may need some redesign.

Answer 18

thank you