Question

Need help with pre-cal review
1. Write the equation of a parabola with vertex (1, -9) and directrix y = -16.

2. What is the focus of the equation that you found in problem #1?

3. What is the axis of symmetry of the equation you found in problem #1?

Answer 1

i swear your book has the formula's to find each of these

Answer 2

Vertex form of a parabola is
Y=a(x-h)^2+k
h and k are the vertex
So we only need to solve for a

Knowing the directrix is -16 and the y value of the vertex is -9, we find the difference of those to be 7, so we add 7 to -9 and that gives us the value of a.

^_^

Answer 3

So would it be -9= -2(x-1)^2 + -9?

Answer 4

Ya, but replace -9= with y=

Answer 5

so the equation is y=-2(x-1)^2+(-9)?

Answer 6

Ya, but u don't need the () around the -9

Answer 7

I was just making sure the negative was obvious. The format in my text book is different though. The equations are set up like (y-k)^2=4p(x-h).

Answer 8

So it'd be like (y+9)^2=4p(x+1). What is "p" though?

Answer 9

For part 2, remember the -2 u got for a? That is the y value for the focus. And the x value for the focus is the x value from the vertex

Answer 10

I think their p is the same as our a

Answer 11

P is the distance between the focus and vertex

Answer 12

That equation from the book looks like it's for parabolas that open sideways not vertical...

Answer 13

Yup, they're sideways.

Answer 14

Yayaya, then p is the same as our a

Answer 15

So (y+9)^2=-8(x-1)?

Answer 16

If your equation for Our parabola given supposed to be open sideways then our directrix should be x equals not Y equals

Answer 17

Yes that must be correct

Answer 18

No, the book says if it opens upward (vertical) then it'd be x equals

Answer 19

Alright then. That's different then what I learned but u do what looks right

Answer 20

Haha. Thanks for your help. :)

Answer 21

^_^