### Instruction: The graph of a parabola has the following properties:

$\bullet$ It passes through the point $(1,5).$

$\bullet$ The $y$-coordinate of the focus is 3.

$\bullet$ Its axis of symmetry is parallel to the $x$-axis.

$\bullet$ Its vertex lies on the $y$-axis.

Express the equation of the parabola in the form
\[ax^2 + bxy + cy^2 + dx + ey + f = 0,\]where $a,$ $b,$ $c,$ $d,$ $e,$ $f$ are integers, $c$ is a positive integer, and $\gcd(|a|,|b|,|c|,|d|,|e|,|f|) = 1.$.### Response: I know that the equation of a parabola with a horizontal axis of symmetry and a vertex at $(h,k)$ is of the form $(y - k)^2 = 4p(x - h),$ where $p$ is the distance from the vertex to the focus and the directrix.
Since the vertex lies on the $y$-axis, I can assume that $h = 0$ and write $(y - k)^2 = 4p(x - 0).$
I also know that the focus is at $(h + p, k),$ so in this case, it is at $(p, k).$
The problem says that the $y$-coordinate of the focus is 3, so I can write $k = 3.$
Now I have $(y - 3)^2 = 4px.$
To find $p,$ I can use the fact that the parabola passes through the point $(1,5).$
Plugging in $x = 1$ and $y = 5,$ I get $(5 - 3)^2 = 4p(1),$ which simplifies to $4 = 4p.$
Therefore, $p = 1.$
Now I have the equation of the parabola in the form $(y - 3)^2 = 4x.$
To express it in the standard form, I need to expand the square, subtract 4x from both sides, and rearrange the terms.
I get $y^2 - 6y + 9 - 4x = 0.$
This is already in the form $ax^2 + bxy + cy^2 + dx + ey + f = 0,$ where $a = 0,$ $b = 0,$ $c = 1,$ $d = -4,$ $e = -6,$ and $f = 9.$
The coefficients are integers, $c$ is positive, and the greatest common divisor of their absolute values is 1.
Therefore, this is the final answer.