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

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- Jul 6th 2006, 09:15 AMbobby77please very urgent.....
the focus of the parabola 4y^2 + 12x - 12y + 39 =0

- Jul 6th 2006, 09:35 AMSoroban
Hello, bobby77!

Quote:

The focus of the parabola: $\displaystyle 4y^2 + 12x - 12y + 39 \:=\:0$

We must the equation into "standard form".

. . . $\displaystyle 4y^2 - 12y \;= \;-12x - 39$

Divide by 4: .$\displaystyle y^2 - 3y \;= \; -3x - \frac{39}{4}$

Complete the square: .$\displaystyle y^2 - 3y + \frac{9}{4} \;= \; -3x - \frac{39}{4} + \frac{9}{4}$

and we have: .$\displaystyle \left(y - \frac{3}{2}\right)^2\;=\;-3x -\frac{15}{2}$

Factor: .$\displaystyle \left(y - \frac{3}{2}\right)^2\;=\;-3\left(x + \frac{5}{2}\right) $*. . . standard form!*

This parabola opens to the left: $\displaystyle \supset$

Its vertex is: $\displaystyle \left(-\frac{5}{2},\,\frac{3}{2}\right)$

Since $\displaystyle 4p = -3$, we have: $\displaystyle p = -\frac{3}{4}$

The focus is $\displaystyle \frac{3}{4}$ units to the__left__of the vertex.

Therefore, the focus is: .$\displaystyle \left(-\frac{13}{4},\,\frac{3}{2}\right)$

- Jul 6th 2006, 01:08 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. - Jul 6th 2006, 01:23 PMCaptainBlackQuote:

Originally Posted by**Quick**

RonL - Jul 6th 2006, 01:43 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". - Jul 6th 2006, 01:47 PMSoroban
Hello, Quick!

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

The equation is: .$\displaystyle 4y^2 + 12x - 12y + 39 \:=\:0$

The graph is comprised of**all**points $\displaystyle (x,y)$ which satisfy the equation.

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

The equation of a circle is: .$\displaystyle x^2 + y^2\:=\:9$ . . . this represents the*entire*circle.

If they wanted only the upper semicircle, they must add: $\displaystyle \text{ for }y \geq 0$

. . or write: .$\displaystyle y \:= \:\sqrt{9 - x^2}$ which, by convention, is the positive square root.

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

. . This is not true . . .

- Jul 6th 2006, 01:54 PMQuickQuote:

Originally Posted by**CaptainBlack**

still, what is the definition of a focus?