Re: Focus of a parabola
Hello, confused0874!
Find the focus of the parabola \(\displaystyle y^2\,=\,12x\)
There are some general rules for "eyeballing" the equation . . .
\(\displaystyle x^2\,=\,4py\:\) opens right or left: \(\displaystyle \,\subset\,\) or \(\displaystyle \,\supset\)
\(\displaystyle y^2\,=\,4px\;\) opems up or down: \(\displaystyle \,\cup\,\) or \(\displaystyle \,\cap\)
If \(\displaystyle 4p\) is positive, the parabola opens in the "positve" direction (right or up).
If \(\displaystyle 4p\) is negative, the parabola opens in the "negative" direction (left or down).
\(\displaystyle p\) is the directed distance from the vertex to the focus.
\(\displaystyle \;\;\)Note: the focus is always "inside the bend" of the parabola.
Your problem: \(\displaystyle \,y^2\,=\,12x\)
The vertex is at the origin: \(\displaystyle (0,0)\)
Since the \(\displaystyle y\) is squared, it opens sideways (right or left).
Since the "4p" is
+12, it opens to the right: \(\displaystyle \,\subset\)
Code:
. . . - | - - * - -
. . . - | -*- - - -
. . . - |*- - - - -
. . . - | - - - - -
. . . - * - - - o <-- focus
. . . - | - - - - -
. . . - |*- - - - -
. . . - | -*- - - -
. . . - | - - * - -
Since \(\displaystyle 4p\,=\,12\;\;\Rightarrow\;\;p\,=\,3\)
The focus is 3 units to the right of the vertex: \(\displaystyle \,F(3,0)\)