Axis of a parabola is y=x and vertex and focus are at a distance 2...

As distance of vertex from origin is $\sqrt{2}$ and focus is $2 \sqrt{2}$. $\therefore V(1,1)$ and F(2,2) [i.e. lying on $y=x$ ] where, length of latusrectum $$\begin{aligned}& =4 a, \\ & =4 \sqrt{2} \end{aligned}$$[\because \text{ where }$a=$\sqrt{2}]\therefore$By definition of parabola P M²=(4 a)(P N) where,$P N$is length of perpendicular upon$x+y-2=0$[i.e. tangent at vertex]$$\begin{array}{ll} \Rightarrow \frac{(x-y)^2}{2}=4 \sqrt{2} \left(\frac{x+y-2}{\sqrt{2}}\right)\\ \therefore & (x-y)^2=8(x+y-2) \end{array}$$ Hence, the equation of parabola is(x-y)^2=8(x+y-2).