Finding the focus of a parabola.

[Guest]

I don't know how to solve this problem.

What is the focus of a parabola whose equation is \(\displaystyle y^2 - 12y - x + 8 = 0\)?

A (28 1/4, 6)
B (-28, 5 3/4)
C (-28, 6 1/4)
D (-27 3/4, 6)


To start with, I did this:

\(\displaystyle y^2 - 12y = x - 8\)

\(\displaystyle y^2 - 12y + 36 = x - 8 + 36\)

\(\displaystyle (y - 6)^2 = x + 28\)

So the vertex is (28, -6) but I don't know what to do after that.

Please help!

Thanks in advance!

 

[stapel]

They should have given you the focus form of the quadratic:

For an upward- or downward-opening parabola with the following equation:

. . . . .(y - k)² = 4p(x - h)

...the vertex is at (h, k) and the focus is at (h + p, k). (The sign on "p" will be negative if the parabola opens downward.)

Hint: Your equation rearranges as (y - 6)² = 4(1/4)(x + 28).

Eliz.

 

[Guest]

They should have given you the focus form of the quadratic:

For an upward- or downward-opening parabola with the following equation:

. . . . .(y - k)² = 4p(x - h)

...the vertex is at (h, k) and the focus is at (h + p, k). (The sign on "p" will be negative if the parabola opens downward.)

Hint: Your equation rearranges as (y - 6)² = 4(1/4)(x + 28).

Eliz.

Okay. I always have the hardest time understanding math.


Where does the 4 in front of 4p come in? I've never seen that before in a general equation.

And also, when the equation rearranges, where does the "4(1/4)" come in? That ends up as 1, so why not just write it as 1? Or is there something I need the 4 or 1/4 to do first?

 

[stapel]

Where does the 4 in front of 4p come in?

Dunno. Something to do with the geometry.

I've never seen that before in a general equation.

Perhaps your book used a different variable when it gave you a formula for finding the focus...?

That ends up as 1, so why not just write it as 1?

Look at the formula. Note what you need. Now read off what you need from the rearranged equation.

Eliz.

 

[Guest]

Where does the 4 in front of 4p come in?

Dunno. Something to do with the geometry.

I've never seen that before in a general equation.

Perhaps your book used a different variable when it gave you a formula for finding the focus...?

That ends up as 1, so why not just write it as 1?

Look at the formula. Note what you need. Now read off what you need from the rearranged equation.

Eliz.

Okay, well. My book says "a" in place of your 4p. It also says that \(\displaystyle a = 1/4c\)

if a=1/4c and a is 1 then we set those equal so it's

\(\displaystyle 1=1/4c\)

then c = 4


I'm pretty sure the parabola opens left or right, so the equation for the focus is (h + c, k). If I'm right, then the focus is (32. -6) but I can't be right, because that's not an option in the answer choices.

I don't know what to do with this. Am I forgetting a valuable step or what?

 

[Guest]

err no, "a" is not in place of your 4p, but the general equation in my book is written as follows:

x - h = a(y - k)₂, where a = 1/4c


and continues to say

vertex(h, k)
focus(h + c, k)
directrix: x = h - c