### 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 with an axis of symmetry parallel to the $x$-axis has an equation of 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 and the directrix.
Since the vertex lies on the $y$-axis, $h = 0$ and the equation simplifies to $(y-k)^2 = 4px.$
To find $k,$ I can use the fact that the focus has a $y$-coordinate of 3.
The focus is the point where the parabola is closest to the axis of symmetry, so its $x$-coordinate must be $p.$
Therefore, the focus is $(p,3)$ and $k = 3.$
To find $p,$ I can use the fact that the parabola passes through the point $(1,5).$
Substituting $x = 1$ and $y = 5$ into the equation, I get $(5-3)^2 = 4p(1),$ or $4 = 4p.$
So $p = 1.$
Now I have the equation $(y-3)^2 = 4x.$
To express it in the desired form, I need to expand the square and move everything to one side.
I get $y^2 - 6y + 9 - 4x = 0.$
Comparing this with $ax^2 + bxy + cy^2 + dx + ey + f = 0,$ I can see that $a = 0,$ $b = 0,$ $c = 1,$ $d = -4,$ $e = -6,$ and $f = 9.$
I can also see that $\gcd(|a|,|b|,|c|,|d|,|e|,|f|) = 1,$ since none of the coefficients have any common factors.
Therefore, the equation of the parabola is $y^2 - 6y - 4x + 9 = 0.$