focus of parabola

Printable View

July 6th 2006, 09:15 AM
bobby77

please very urgent.....

the focus of the parabola 4y^2 + 12x - 12y + 39 =0
July 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: . $\left(-\frac{13}{4},\,\frac{3}{2}\right)$
July 6th 2006, 01:08 PM
Quick

1 Attachment(s)

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.
July 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
July 6th 2006, 01:43 PM
ThePerfectHacker

1 Attachment(s)

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, 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 . . .
July 6th 2006, 01:54 PM
Quick

1 Attachment(s)

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?