1. ## parabola

In need to write an equation for the parabola whose vertex is (4, 3), axis of symmetry is y = 3 and the measure of the latus rectum is 4; a > 0. I also need to draw this parabola.

2. A parabola is the set of all points in the plane equidistant from a given line.
$\displaystyle \sqrt{(y-12)^2+(x-3)^2}=y+4$

3. Hello, 14041471!

Write an equation for the parabola whose vertex is (4, 3),
axis of symmetry is $\displaystyle y = 3$ and the measure of the latus rectum is 4; $\displaystyle a > 0.$
I also need to draw this parabola.

Code:
        |
|                 *
|             *
|          *  :
|  (4,3) *  p :
- + - - - o - - o - -
|      V *    :F
---+----------*--:------
|             *
|                 *
|
We're expected to know the following . . .

Since the axis of symmtetry is horizontal, the parabola opens "sideways."
Since $\displaystyle a > 0$, the parabola opens to the right.

The general form is: .$\displaystyle (y-k)^2 \:=\:4p(x-h)$
. . where $\displaystyle (h,k)$ is the vertex,
. . and $\displaystyle p$ is the focal distance (distance from vertex to focus).

Fact: the length of the latus rectum is $\displaystyle 4p.$
. . . . We have: .$\displaystyle 4p \:=\:4\quad\Rightarrow\quad p \:=\:1$

Since the vertex is: .$\displaystyle (h,k) = (4,3)$

. . the equation is: .$\displaystyle (y-3)^2 \;=\;4(x-4)$

4. ! Wrong direction for me. May I swap x's and y's?