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.
You might start with a sketch . . . 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: .$\displaystyle (yk)^2 \:=\:4p(xh)$
. . 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: .$\displaystyle (y3)^2 \;=\;4(x4)$