### 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 recall that the standard form of a parabola with an axis of symmetry parallel to the $x$-axis is $(y - k)^2 = 4p(x - h),$ where $(h,k)$ is the vertex and $p$ is the distance from the vertex to the focus.
I also remember that the focus of such a parabola is $(h + p, k),$ so the $y$-coordinate of the focus is $k.$
Since the vertex lies on the $y$-axis, I know that $h = 0,$ so the vertex is $(0,k)$ and the focus is $(p,k).$
The problem gives me that the $y$-coordinate of the focus is 3, so $k = 3.$
Therefore, the vertex is $(0,3)$ and the focus is $(p,3).$
To find $p,$ I can use the fact that the parabola passes through the point $(1,5).$
Plugging in $x = 1$ and $y = 5$ into the standard form, I get $(5 - 3)^2 = 4p(1 - 0),$ which simplifies to $4 = 4p.$
Solving for $p,$ I get $p = 1.$
So, the focus is $(1,3).$
Now, I can write the standard form of the parabola as $(y - 3)^2 = 4(x - 0),$ which simplifies to $y^2 - 6y + 9 = 4x.$
To express this equation in the required form, I need to subtract $4x$ from both sides and rearrange the terms.
I get $-4x + y^2 - 6y + 9 = 0.$
This is already in the form $ax^2 + bxy + cy^2 + dx + ey + f = 0,$ where $a = -4,$ $b = 0,$ $c = 1,$ $d = 0,$ $e = -6,$ and $f = 9.$
I notice that the coefficients are all integers, $c$ is a positive integer, and the greatest common divisor of the absolute values of the coefficients is 1, as required.
Therefore, this is the final answer.