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

2. Hello, bobby77!

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

3. ## What is this?!?

Originally Posted by Soroban
This parabola opens to the left: $\displaystyle \supset$
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" graph program.

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

RonL

5. 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".

6. 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 . . .

7. 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?