the focus of the parabola 4y^2 + 12x - 12y + 39 =0

Printable View

- July 6th 2006, 10:15 AMbobby77please very urgent.....
the focus of the parabola 4y^2 + 12x - 12y + 39 =0

- July 6th 2006, 10:35 AMSoroban
Hello, bobby77!

Quote:

The focus of the parabola:

We must the equation into "standard form".

. . .

Divide by 4: .

Complete the square: .

and we have: .

Factor: .*. . . standard form!*

This parabola opens to the left:

Its vertex is:

Since , we have:

The focus is units to the__left__of the vertex.

Therefore, the focus is: .

- July 6th 2006, 02:08 PMQuickWhat is this?!?Quote:

Originally Posted by**Soroban**

**What's going on!?!**

P.S. I would like to thank hacker for the great "trojan" :D graph program. - July 6th 2006, 02:23 PMCaptainBlackQuote:

Originally Posted by**Quick**

RonL - July 6th 2006, 02:43 PMThePerfectHacker
Quick you are using the program I use nice job.

There is a smarther thing you can do (you flawed as CaptainBlack said).

Click on the button indicated and type the regular equation without the need to solve for "y". - July 6th 2006, 02:47 PMSoroban
Hello, Quick!

Why are you considering only the upper half of the parabola?

The equation is: .

The graph is comprised of**all**points which satisfy the equation.

If they wanted only the upper half, they would have to say so.

The equation of a circle is: . . . . this represents the*entire*circle.

If they wanted only the upper semicircle, they must add:

. . or write: . which, by convention, is the positive square root.

You may be thinking that all equations must represent*functions*(single-valued).

. . This is not true . . .

- July 6th 2006, 02:54 PMQuickQuote:

Originally Posted by**CaptainBlack**

still, what is the definition of a focus?