Question

Identify the vertex, focus, and directrix of the parabola with the equation x^2 – 6x – 8y + 49 = 0

Answer 1

@thomaster ?

Answer 2

Do you know how to complete the square?

Answer 3

no..could you please show me?

Answer 4

Okay, \[x^2-6x+n\]is an expression which could be written like \[(x+a)^2\]if we only knew what value to use for \(a\), right?

\[(x+a)^2 = (x+a)(x+a) = x^2 + ax + ax + a^2 \]\[\qquad=x^2+2ax+a^2\]

If we line those two up over each other:
\[x^2-6x+n\]\[x^2+2ax+a^2\]

They will be equal if (and only if)\[-6x=2ax\]and\[n=a^2\]

Answer 5

umm..ok, them what do we need to do?

Answer 6

so, our job is to figure out how much to take from that 49 we have sitting there in the equation so that we can write our equation as something like \[(x+a)^2 = 8y+b\]

So, \[-6x=2ax\]what does \(a=\)

Answer 7

umm 2ax? im not sure..how do you find a?

Answer 8

\[-6x=2ax\]Solve for \(a\)
Do you see any common factors you can cancel?

Answer 9

x?

Answer 10

ok, so -6=2a?

Answer 11

Yes, good so far, what is the value of \(a\) if that is the case?

Answer 12

im not sure, how would you find it? we know that 2a=-6...?

Answer 13

i just dont know what to do next..

Answer 14

Uh, to solve for \(a\) in that equation, wouldn't you divide both sides by the number that appears in front of the \(a\)?

Answer 15

What number times 2 gives you -6? That number is \(a\).

Answer 16

ooohh yeah! ok, so a would equal 3

Answer 17

wow sorry blond moment:P

Answer 18

okay, so we have \(a = -3\)

what is \(a^2=\)

Answer 19

oh i forgot about the negative... so it would be -3

Answer 20

a^2=9

Answer 21

then what?

Answer 22

Ok, so we've found that we can take \(x^2-6x\) and add \(9\) to it to get a "perfect square" which we can write as \((x-3)^2\):

\[(x-3)^2=(x-3)(x-3) = x(x-3)-3(x-3) = x*x-3x-3x-3(-3)\]\[\qquad = x^2-6x+9\]

Any question about that?

Answer 23

(other than "why would you want do to THAT?!?" :-)

Answer 24

haha:) umm so you just expand it in order to simplify it?

Answer 25

We actually want to keep in the compact form we have: \((x+3)^2\)

But let's take a minute and see just what we did:

We had \[x^2-6x+\text{<stuff not involving x>}\]
We found that if we took half of the number in front of \(x\) (call it \(a\)), and squared it (\(a^2\)), we could write \((x+a)^2\) instead of \(x^2-6x\)

However, we need to make an adjustment. We're taking \(a^2=9\) away from the rest of the equation, so we can't just write\[(x-3)^2-8y+49=0\]and expect good things to happen. Instead, we need to balance the equation so that it is still a valid equation.

To write \((x-3)^2\) we needed another \(a^2=9\). If we just add that value to both sides of the equation, then we're in the clear, because you can add the same quantity to both sides of an equation without disturbing the balance.

\[x^2-6x -8y + 49 = 0\]\[x^2-6x +9 -8y + 49 = 0+9\]\[(x^2-6x+9)-8y+49=9\]\[(x-3)^2-8y+49=9\]Do those steps make sense to you?

Answer 26

Think of it as making neat square arrays of blocks. You've got some blocks already on the table, but you need a few more to complete your square. The teacher wants to keep things fair, so that you don't get more blocks than the kid next to you, so when you raise your hand and say "I need 9 more blocks to complete my square", the teacher comes over and gives you 9 more blocks, and gives the kid next to you 9 more blocks. You've still both got the same number of blocks as each other, but now you can arrange yours in a square.

Answer 27

oh, ok:) well what do we do next?

Answer 28

or are we done?

Answer 29

So, we've completed the square. Our work is just beginning :-)

Answer 30

ooohh..yey..:)

Answer 31

what forms of the parabola equation have you learned so far?

Answer 32

have you heard of vertex form?

Answer 33

yes, that and standard form..

Answer 34

Okay, does what we have here:
\[(x-3)^2 -8y + 49=9\]look like it would be easy to put into vertex form?

Answer 35

well vertex form is y=a(x-h)^2+k...and we dont know what (x-h)^2 would be..

Answer 36

Why don't you try solving what we have for \(y\)

Answer 37

its -8y

Answer 38

Take the equation we have, and use algebra to rearrange it so we have \[y=\text{<stuff>}\]

Answer 39

\(y\) will not appear in <stuff>
the only \(y\) will be all alone on the left side

Answer 40

\[(x-3)^2-8y+49=9\]Add \(8y\) to both sides\[(x-3)^2-8y+49+8y=9+8y\]\[(x-3)^2+49=9+8y\]Any question about that?

Answer 41

Next, flop the equation:
\[9+8y=(x-3)^2+49\]
Any question about that?

Answer 42

no questions:) whats next?:)

Answer 43

What's next is I would like you to complete the process of isolating \(y\)

Answer 44

You're going to have to subtract 9 from both sides, and then divide both sides by some number

Answer 45

ok, so we would have 8y-(x-3)^2+40...and now we divide by 8 on each side right?

Answer 46

what happened to the = sign?

Answer 47

oh, whoops..8y=(x-3)^2+40

Answer 48

that's better.

Now divide both sides by 8

Answer 49

ok...y=(x-3)^2+5

Answer 50

yes, does that look like vertex form?

Answer 51

yes:)

Answer 52

okay, so what is the vertex?

Answer 53

(x-3)^2

Answer 54

right?

Answer 55

no. the vertex is a point

Answer 56

vertex form:
\[y = a(x-h)^2+k\]vertex is at \((h,k)\)

Answer 57

No!

Answer 58

Compare equations:

\[y = a(x-h)^2+k\]\[y=(x-3)^2+5\]

Answer 59

what is the value of \(k\)?

Answer 60

im not sure?

Answer 61

oh! its 5..

Answer 62

Yes, \(k=5\). What is the value of \(h\)?

Answer 63

Well, it appears you left sometime during the time that OpenStudy was taking its usual mid-morning nap.

The vertex is at \((h,k)\) once you figure out the value of \(h\). Note very carefully that the formula is \[y = a(x-h)^2+k\]so make sure you get the proper sign!

Answer 64

We have a parabola with \(x^2\) in it, so it opens either up or down. A parabola with \(y^2\) in it would open either right or left.

If the parabola opens up or down, the directrix will be a line with a constant \(y\) value. The directrix and the focus are located equidistant from the vertex.