Question

Find the vertex, focus, and directrix of the parabola with the equation x^2-6x-8y+49=0
I know the answers, I just need to figure out how you get them

Answer 1

To start, have you tried to put the equation into vertex form of a parabola?

Answer 2

no, how do I put it into the vertex form? is there an equation for that?

Answer 3

i know something like a(x-h)^2+k

Answer 4

We would have to complete the square with the quadratic part first. That will get us the (x - h)^2 part. Then solve for y.
\( \color{blue}{x^2 - 6x} - 8y + 49 = 0 \)

Answer 5

what exactly is compleating the square?

Answer 6

im sorry that you have to explain everything, i do apprishate it

Answer 7

The process is to take the coefficient on the linear term, -6x, divide it by two and then square it. We would add the new number onto both sides of the equation OR add and subtract that number on the same side of the equation (either way does not change the equation, only the way it is written).

The theory goes into how we expand the binomial squared: \( (x-h)^2 = x^2 - 2h x + h^2 \)
We have the \(x^2 - 2h x\) part. The part we add on is the linear term, \(-2h\), divided by two and then squared: \((-2h2)^2 = h^2\)
http://www.regentsprep.org/Regents/math/algtrig/ATE12/completesqlesson.htm
This might be a helpful reminder as well, since I don't know how much justice I do the explanation of that one!

Answer 8

And some other examples:
x^2 - 4x. We add on ( -4 /2)^2 = 2^2 = 4 to complete the square. x^2 - 4x + 4 - 4 = (x - 2)^2 - 4
x^2 + 10x. We add on ( 10/2)^2 = 5^2 = 25 to complete the square. x^2 + 10x + 25 - 25 = (x + 5)^2 - 25.

Answer 9

so -6x = -3x = 9x ?

Answer 10

x^2 -6x -8y + 49=0
x^2 -8y + 49= 9

Answer 11

thennn : x^2 -8y= -49 + 9
?

Answer 12

We won't include the x part. We only use the coefficient itself. But I agree with the number.
(-6/2)^2 = 9 is what we have to add on to complete the square.
\( x^2 - 6x - 8y + 49 = 0 \)
\( x^2 - 6x \color{green}{+ 9} - 8y + 49 = \color{green}{+9} \)
The x^2 - 6x + 9 is now a perfect square trinomial, which factored gives us (x - 3)^2.
\( (x - 3)^2 - 8y + 49 = 9 \)

Answer 13

okay i see how you did that, what next?

Answer 14

do i factor?

Answer 15

oh wait the 49 needs to go on the other side right?

Answer 16

then i factor after i subtract the 49

Answer 17

then it will look like
(x-3) -8y = 9 - 49
(x-3) -8y = -40

Answer 18

wait i did that wrong

Answer 19

i have confused myself

Answer 20

Our next goal would then be solving for y, to put it in vertex form. So we can indeed combine those constants together. I think it would be better to subtract 9 from both sides and add 8y to both sides.
(x-3)^2 + 40 = 8y
Then we just need to divide both sides by the coefficient on y to have y alone.

Answer 21

After we completed the square, our goal was to solve for y. We want the equation put into vertex form so that we can first find the vertex, and then find focus/directrix.

Answer 22

okay so

(x-3)^2 + 40 = 8y
(x-3)^2 = 8y - 40
(x-3)^2 = 8(y-5)
y=3, y=5

Answer 23

is that what you mean?

Answer 24

We don't need to do anything like that:
Solving for y means to get y alone on one side of the eqn. At that first line, all we gotta do is divide by 8 from both sides to get y alone!
\( \dfrac{1}{8} \left[ (x-3)^2 + 40 \right] = \dfrac{1}{8} \left( 8y \right) \)
\( \dfrac{1}{8} (x - 3)^2 + 5 = y \)
Can you see how this gets our vertex equation now?

Answer 25

okay let me write that down

Answer 26

where did the 9 go?

Answer 27

Sure. And I can do a summary post at the end just to make sure its all clear.

Answer 28

wait just kidding

Answer 29

Alright. Any other issues so far? It all makes sense up to this point?

Answer 30

crystal clear, so what next?

Answer 31

The vertex form actually gives us the vertex for free.
\( y = a(x - h)^2 + k \)
The vertex is the point (h, k).

Our equation is: \( y = \dfrac{1}{8} (x - 3)^2 + 5 \)
Can you pick out the values here for the vertex?

Answer 32

yes, vertex (3,5) ?

Answer 33

Looks good!

Focus and directrix are a little trickier. The way I've been explaining it is, first we need to find the distance from vertex to focus.
This is found with the a-value, which in our case is 1/8. a = 1/4c, where c is the length between focus and vertex.
So we first solve this equation: 1/8 = 1/(4c) for c.

Answer 34

okay i understand, but where did the 4 come from? are they all always 4 or is the 4 being used because it is half of 8?

Answer 35

It is always 1/(4c) = a.
This reference might help. http://home.windstream.net/okrebs/page64.html .
And between y = ax^2 and y = a(x - h)^2 + k, they have the same distance between focus and vertex. The translation does not affect it. :)

Answer 36

oh okay, so how do we get the focus from that?

Answer 37

The distance from the vertex to the focus... so when we have the distance, we are simply shifting the vertex by that distance onto the place where the focus is. Like this:

This requires a bit of subjective understanding as well. This is where the parabola is facing upwards only. If it faces downwards, the focus is below the vertex by the same distance.