# focus of parabola

• Jul 6th 2006, 09:15 AM
bobby77
the focus of the parabola 4y^2 + 12x - 12y + 39 =0
• Jul 6th 2006, 09:35 AM
Soroban
Hello, bobby77!

Quote:

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

We must the equation into "standard form".

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

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

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

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

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

This parabola opens to the left: $\supset$

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

Since $4p = -3$, we have: $p = -\frac{3}{4}$
The focus is $\frac{3}{4}$ units to the left of the vertex.

Therefore, the focus is: . $\left(-\frac{13}{4},\,\frac{3}{2}\right)$
• Jul 6th 2006, 01:08 PM
Quick
What is this?!?
Quote:

Originally Posted by Soroban
This parabola opens to the left: $\supset$

I only just got through algebra1 this year, but the graph doesn't show a parabola, it shows half of one. Considering it only shows half of one, there is no line of symmetry, is the point it starts considered a vertex? What is the focus? What's going on!?!

P.S. I would like to thank hacker for the great "trojan" :D graph program.
• Jul 6th 2006, 01:23 PM
CaptainBlack
Quote:

Originally Posted by Quick
I only just got through algebra1 this year, but the graph doesn't show a parabola, it shows half of one. Considering it only shows half of one, there is no line of symmetry, is the point it starts considered a vertex? What is the focus? What's going on!?!

P.S. I would like to thank hacker for the great "trojan" :D graph program.

You might want to try plotting the other branch: f(x)=1.5-sqrt(....)

RonL
• Jul 6th 2006, 01:43 PM
ThePerfectHacker
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 PM
Soroban
Hello, Quick!

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

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

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

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

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

If they wanted only the upper semicircle, they must add: $\text{ for }y \geq 0$
. . or write: . $y \:= \:\sqrt{9 - x^2}$ 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 PM
Quick
Quote:

Originally Posted by CaptainBlack
You might want to try plotting the other branch: f(x)=1.5-sqrt(....)

RonL

Oops, thanx.

still, what is the definition of a focus?