the focus of the parabola 4y^2 + 12x - 12y + 39 =0
The focus of the parabola:
We must the equation into "standard form".
. . .
Divide by 4: .
Complete the square: .
and we have: .
Factor: . . . . standard form!
This parabola opens to the left:
Its vertex is:
Since , we have:
The focus is units to the left of the vertex.
Therefore, the focus is: .
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.
Why are you considering only the upper half of the parabola?
The equation is: .
The graph is comprised of all points which satisfy the equation.
If they wanted only the upper half, they would have to say so.
The equation of a circle is: . . . . this represents the entire circle.
If they wanted only the upper semicircle, they must add:
. . or write: . which, by convention, is the positive square root.
You may be thinking that all equations must represent functions (single-valued).
. . This is not true . . .