### 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 a parabola whose axis of symmetry is parallel to the $x$-axis has the form $(y-k)^2 = 4p(x-h),$ where $(h,k)$ is the vertex and $p$ is the distance from the vertex to the focus.
Since the vertex lies on the $y$-axis, I can assume that $h = 0.$
Then the equation becomes $(y-k)^2 = 4px.$
I also know that the focus is at $(h+p,k),$ so if the $y$-coordinate of the focus is 3, then $k = 3.$
Then the equation becomes $(y-3)^2 = 4px.$
I can use the given point $(1,5)$ to find $p.$
Substituting $x = 1$ and $y = 5,$ I get $(5-3)^2 = 4p(1),$ which simplifies to $4 = 4p.$
Dividing by 4, I get $p = 1.$
Then the equation becomes $(y-3)^2 = 4x.$
To express it in the desired form, I need to expand the square and rearrange the terms.
I get $(y-3)^2 = 4x \implies y^2 - 6y + 9 = 4x \implies -4x + y^2 - 6y + 9 = 0.$