the focus of the parabola 4y^2 + 12x - 12y + 39 =0
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: .$\displaystyle \left(-\frac{13}{4},\,\frac{3}{2}\right)$
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!?!Originally Posted by Soroban
P.S. I would like to thank hacker for the great "trojan" graph program.
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 . . .