Is This Just Another Boring Parabola Problem?

First we find the slope of the equation that passes through $V$ and $F$ , which is called the focal axis:

$m=\dfrac{6-3}{5-2}=1$

Now, we will find the directrix, but to do that we need to find a new point $Q$ , such that $V$ is the midpoint of $\overline{FQ}$ , so we will apply the formula of a midpoint with $Q=(a,b)$ :

$2=\dfrac{5+a}{2} \Longrightarrow a=-1$

$3=\dfrac{6+b}{2} \Longrightarrow b=0$

Hence, $Q=(-1,0)$ . Now, we also need to find the slope of the perpendicular line to the focal axis, that is: $m'=-\dfrac{1}{m}=-\dfrac{1}{1}=-1$ .

With that, now we can find the directrix, knowing its slope ( $m'$ ) and a point ( $Q$ ):

$y-0=-1(x+1) \Longrightarrow x+y+1=0$

Finally, we find our parabola (points $P=(x,y)$ ) applying its definition:

$\overline{FP}=$ distance from $x+y+1=0$ to $P$

$\sqrt{(x-5)^2+(y-6)^2}=\dfrac{|x+y+1|}{\sqrt{2}}$

Square both sides, expand and simplify:

$(x-5)^2+(y-6)^2=\dfrac{(x+y+1)^2}{2}$

$2(x^2-10x+25+y^2-12y+36)=x^2+y^2+1+2xy+2x+2y$

$x^2-2xy+y^2-22x-26y+121=0$

Hence, $a=1$ , $b=-2$ , $c=1$ , $d=-22$ , $e=-26$ , $f=121$ and $|a+b+c+d+e+f|=73$ .

Distance between focus and vertex is $3\sqrt{2}$ , so if the parabola were canonical its equation would have been $y^2=12\sqrt{2} x$ . The slope of the line between the vertex and the focus is 1, so in order to obtain this parabola from the canonical one we need to rotate by 45 degrees, then translate. Let $\begin{bmatrix} u \\ v \\ \end{bmatrix}= \begin{bmatrix} 2 \\ 3 \\ \end{bmatrix} + \begin{bmatrix} \frac{\sqrt2}{2} & -\frac{\sqrt2}{2}\\ \frac{\sqrt2}{2} & \frac{\sqrt2}{2} \\ \end{bmatrix} \begin{bmatrix} x \\ y \\ \end{bmatrix}$ . A simple calculation shows: $u+v=5+\sqrt2 x$ , $u-v=-1-\sqrt2 y$ , So that $y^2=\frac{(v-u-1)^2}{2}$ and $x=\frac{u+v-5}{\sqrt2}$ . Substitute in the original equation to get $v^2-2uv+u^2-26v-22u+121=0$ .