JEE Main 2025 — 4 April, Evening Shift — Question 26

The vertex and focus lie on the axis $y=x$. The distance from the origin to the vertex is $\sqrt{2}$, so vertex is at $(1,1)$. The distance from the origin to the focus is $2\sqrt{2}$, so focus is at $(2,2)$. The distance from the vertex to the focus is $\sqrt{2}$, so $a = \sqrt{2}$. The directrix is perpendicular to the axis (slope $-1$) and is at a distance $a$ from the vertex on the opposite side of the focus; thus it passes through the origin and has equation $y = -x$. For a point $(1,k)$ on the parabola, its distance to the focus equals its distance to the directrix:

$$\frac{|k+1|}{\sqrt{2}} = \sqrt{(1-2)^2 + (k-2)^2}.$$

Square both sides:

$$\frac{(k+1)^2}{2} = 1 + (k-2)^2.$$

Multiply by 2 and expand:

$$k^2 + 2k + 1 = 2 + 2(k^2 - 4k + 4) = 2k^2 - 8k + 10.$$

Bring all terms to one side:

$$0 = k^2 - 10k + 9 = (k-1)(k-9).$$

Hence $k = 1$ or $k = 9$. Among the options, only $9$ appears, so $k = 9$.