JEE Advanced · 2006 · 09 Apr 2006 · Parabola

The axis is $y=x$, so the vertex is at $(1,1)$ (distance $\sqrt{2}$ from origin).

The focus is at distance $2\sqrt{2}$ from origin along $y=x$, so focus is at $(2,2)$.

The focal length $a=\sqrt{(2-1{)}^2+(2-1{)}^2}=\sqrt{2}$.

The directrix is perpendicular to $y=x$, so it has equation $x+y=k$.

Since the vertex is equidistant from focus and directrix, and vertex is at $(1,1)$, the directrix is $x+y=0$.

The parabola is the locus of points equidistant from focus $(2,2)$ and directrix $x+y=0$:



Squaring and simplifying: $(x-y{)}^2=8(x+y-2)$.