JEE Main · 2025 · 04 Apr 2025 · Shift 2 · Parabola

The vertex lies on $y=x$ at distance $\sqrt{2}$, namely $(1,1)$, and the focus at distance $2\sqrt{2}$, namely $(2,2)$; the directrix is the perpendicular to the axis through the origin, $x+y=0$. For $(1,k)$ on the parabola, $\sqrt{(1-2{)}^2+(k-2{)}^2}=\frac{|1+k|}{\sqrt{2}}$, which simplifies to $k^2-10k+9=0$, so $k=1$ or $k=9$. Since $(1,1)$ is the vertex, the admissible value is $k=9$.