Question

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

Answer 1

So far I have \[8y=(x-3)^{2} +49\]
What do I do next?

Answer 2

Ouch! Something is missing!

Answer 3

Can you help me @mathmate

Answer 4

The regrouping is missing a 9 somewhere.
From
\(x^2 -6x-8y+49=0\)
we get
\(x^2 -6x+49=8y\)
\(8y=x^2 -6x+49\)
\(8y=(x^2 -6x+9) -9 +49\)
\(8y=(x-3)^2 +40\)
\(8(y-5)=(x-3)^2\)
This means that the vertex of the conic has been translated to (3,5).
Can you take it from here?

Answer 5

Why take away 9 from 49 to make it 40?

Answer 6

Because we need to add a nine to complete the square of (x-3), so we need to subtract 9 on the same side of the equal sign in order not to change the value of the expression.

Answer 7

Oooooh :) Thank you for explaining that

Answer 8

Can you complete the rest of the question?
If you want to confirm your answer, just post.

Answer 9

How did you get the 5? \[8(y-5)=(x-3)^{2}\]

Answer 10

Moved the 40 over to the left, and factor with 8.
8x-40=8(x-5)

Answer 11

Sorry, I meant 8y-40=8(y-5)

Answer 12

From the equation above, when written in standard form of a horizontal parabola \[x= \frac{ 1 }{ 4c }(y-k)^{2}+h\] ; Am I going to use the 8 in the equation?

Answer 13

What you have is a parabola with a vertical axis, because y is linear, and x is to the second degree. You cannot arbitrarily change a parabola with a vertical axis to one with a horizontal axis.
The equation \(8(y-5)=(x-3)^2\) means that h=3, k=5, and p=2 in the equation
\((y-k)=\frac{(x-h)^2}{4p}\)
where p is the distance of the focus from the vertex, and it is also the distance of the vertex from the directrix.
(h,k) is the position of the vertex.

Answer 14

P=2 ?

Answer 15

@amistre64 @mathstudent55 @KayleeNeedsHelp can you help me?

Answer 16

Identify the vertex, focus, and directrix of the parabola with the equation

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

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

(x-3)^2-8y+40=0

(x-3)^2 = 8y-40

(x-3)^2 = 8 (y-5)

1/8 (x-3)^2 = (y-5)

Answer 17

vertex should be easily determined, the p parameter seems to be 2 yes: 4(2)=8

Answer 18

the p parameter just tells us how far to go up and down from the vertex to find the focus and directrix

Answer 19

How did you get 2 though or why 2? I don't remember from the equations I know of having the number 2 being multiplied by 4. Though \[a=\frac{ 1 }{ 4c }\]
I just want to know why 4 is being multiplied by 2 to get 8.

Answer 20

the standard equation is y=x^2/(4p) (some books say y=x^2/(4c)
The equation was made in such a way to recognize the distance between the focus and the vertex, and also the distance between the vertex and the directrix.
Note that if the sign of p (or c) is reversed, the parabola is also reversed.
When p>0, the parabola opens upwards.

Answer 21

... A horizontal parabola opens to the right (positive x) if p>0.

Answer 22

where does the 4 come from? its just a side effect of creating the general equation of a parabola to start with.

we define a parabola as all the points (x,y) that are the same distance from a focus point (a,b). and a directrix line (x=x, y=k)

\[\sqrt{(x-a)^2+(y-b)^2}=\sqrt{(x-x)^2+(y-k)^2}\]
\[(x-a)^2+(y-b)^2=(x-x)^2+(y-k)^2\]
\[(x-a)^2+(y-b)^2=(y-k)^2\]
\[(x-a)^2+y^2-2by+b^2=y^2-2ky+k^2\]
\[(x-a)^2+b^2-k^2=2by-2ky\]
\[(x-a)^2+(b^2-k^2)=2y(b-k)\]
now b-k is the distance between the focus and the directrix line, the parameter 'p' is half of (b-k).
\[p=\frac{(b-k)}{2}\]
\[2p=(b-k)\]
\[4p=2(b-k)\]
and this is why we get a generality with a 4p sitting in the mix.
\[(x-a)^2+(b^2-k^2)=4p~y\]