The axis of a parabola is the line $$y = x$$ and its vertex and focus are in the first quadrant at distances $$\sqrt{2}$$ and $$2\sqrt{2}$$ units from the origin, respectively. If the point (1, k) lies on the parabola, then a possible value of k is :

The axis of the parabola is the line $$y = x$$, so the axis makes an angle of $$45^{\circ}$$ with the positive $$x$$-axis and its unit direction vector is $$\left(\dfrac{1}{\sqrt{2}},\,\dfrac{1}{\sqrt{2}}\right)$$.

On this axis, the vertex $$V$$ and the focus $$S$$ are in the first quadrant and are given to be at distances $$\sqrt{2}$$ and $$2\sqrt{2}$$, respectively, from the origin $$O(0,0)$$.

Moving a distance $$d$$ along the axis from the origin gives the point $$\left(\dfrac{d}{\sqrt{2}},\,\dfrac{d}{\sqrt{2}}\right)$$.
  • For $$d = \sqrt{2}$$ (vertex), $$V\left(1,\,1\right)$$.
  • For $$d = 2\sqrt{2}$$ (focus), $$S\left(2,\,2\right)$$.

The distance from the vertex to the focus is
$$p = VS = \sqrt{(2-1)^2 + (2-1)^2} = \sqrt{2}.$$

To write the equation conveniently, rotate the coordinate system through $$45^{\circ}$$ so that the new $$u$$-axis lies along $$y = x$$.
Define the rotated coordinates:
$$u = \dfrac{x + y}{\sqrt{2}}, \quad v = \dfrac{y - x}{\sqrt{2}}.$$

In $$(u,v)$$-coordinates:
Vertex $$V$$ is $$\left(u_0,0\right) = \left(\sqrt{2},\,0\right)$$.
Focus $$S$$ is $$\left(u_0 + p,\,0\right) = \left(\sqrt{2} + \sqrt{2},\,0\right) = \left(2\sqrt{2},\,0\right)$$.

For a parabola opening in the positive $$u$$-direction with parameter $$p$$, the standard form is
$$v^{2} = 4p\left(u - u_0\right).$$

Substituting $$p = \sqrt{2}$$ and $$u_0 = \sqrt{2}$$:
$$v^{2} = 4\sqrt{2}\left(u - \sqrt{2}\right).$$

Re-express in terms of $$x, y$$:
$$v^{2} = \left(\dfrac{y - x}{\sqrt{2}}\right)^{2} = \dfrac{(y - x)^{2}}{2},$$
$$u = \dfrac{x + y}{\sqrt{2}}.$$ Hence
$$\dfrac{(y - x)^{2}}{2} = 4\sqrt{2}\left(\dfrac{x + y}{\sqrt{2}} - \sqrt{2}\right).$$

Simplify the right-hand side:
$$4\sqrt{2}\left(\dfrac{x + y - 2}{\sqrt{2}}\right) = 4(x + y - 2).$$

Multiplying both sides by $$2$$ gives the Cartesian equation of the parabola:
$$\boxed{(y - x)^{2} = 8\,(x + y - 2)}.$$ Now test the point $$P(1,k)$$.

Substitute $$x = 1,\, y = k$$:
$$(k - 1)^{2} = 8\left(1 + k - 2\right) = 8(k - 1).$$

Let $$t = k - 1$$. Then
$$t^{2} = 8t \quad\Longrightarrow\quad t(t - 8) = 0.$$ Thus $$t = 0 \; \text{or} \; t = 8.$

• $$t = 0 $$\Rightarrow$$ k = 1$$, which corresponds to the vertex point. • $$t = 8 $$\Rightarrow$$ k = 9$$.

Among the given options, the admissible value is $$k = 9$$ (Option B).

Therefore, a possible value of $$k$$ is $$\mathbf{9}$$.