geometry
Using the definition of a parabola as the locus of points equidistant from a given point (focus) and a line (directrix).
Here is my work:
$\sqrt{(y+x)^2}=\sqrt{(x-1)^2+(y-1)^2}$
$y^2+2xy+x^2=x^2-2x+1+y^2-2y+1$
$2xy+2y=-2x+2$
$y=\frac{-x+1}{x+1}$
This equation is a hyperbola not a parabola. Any suggestions on what I have done wrong would be appreciated.
(Sketch.) Let $P$ and $Q$ be two points on a parabola, and let $R$ be the point where the respective tangents to the parabola intersect. Let $X$ the midpoint of $PQ$. Then $RX$ is parallel to the axis of symmetry of the parabola (proved below). Draw lines through $P$ and $Q$ parallel to $RX$, and reflect these lines in the respective tangents; the focus $F$ is the intersection of the reflected lines.
To show that $RX$ is parallel to the axis of symmetry: Drop perpendiculars from $P,Q,R$ to the directrix, meeting it at $P',Q',R'$ respectively. As you alluded to, the tangent at $P$ is the perpendicular bisector of the segment $FP'$, and likewise for $Q$ and $FQ'$. So, in $\triangle FP'Q'$, two of the perpendicular bisectors pass through $R$; therefore the third does as well. Since $RR'$ is a line through $R$ and perpendicular to $P'Q'$, it must be the perpendicular bisector, that is, $R'$ is the midpoint of $P'Q'$. By parallels, (the extension of) $RR'$ bisects $PQ$, that is, $RR'$ passes through $X$. So $RX$ is perpendicular to the directrix, as claimed.
Edit: Just for reference, here's what this looks like analytically: The direction of $RX$ is $(2,1)$; reflecting in $RP$ just means exchanging $x$ and $y$ coordinates, so the direction of $PF$ is $(1,2)$, and the line through $P$ in that direction is $2x-y=1$. Reflecting in $RQ$ means negating the $y$-coordinate, so the direction of $QF$ is $(2,-1)$, and the line through $Q$ in that direction is $x+2y=1$. The intersection of these lines is $(\frac35,\frac15)$.
Assume any point $(x,y)$ on the parabola. Given the focal point $(0,0)$ and the directrix $x+2y=10$, we can establish the equal distance from $(x,y)$ to the focus and the directrix as follows
$$\sqrt{x^2+y^2} = \frac{|x+2y-10|}{\sqrt{1^2+2^2}}$$
Simplify to obtain the equation
$$4x^2-4xy+ y^2 +20x+40y-100=0$$
Best Answer
Let $(x,y)$ be a point on the parabola, its distance to the directrix is given by
$$\frac{|x+y|}{\sqrt{1^2+1^2}}$$
So the equation of the parabola is
$$(x-1)^2+(y-1)^2=\frac{(x+y)^2}{2}$$