If (a²,a-2) be a point interior to the region of the parabola y²=2x bounded by the chord joining the points (2,2)and (8,-4)then the set of all possible real values of a, is---(A) (-2,2^1/2) (B) (-3,2) (c)(-2,2^1/2) (D) (-2,-2+2^1/2)

Below is the graph of the parabola and the chord. The shaded region is the region bound by the parabola $y^2=2x$ and $y-2=\left(\frac{-2-2}{8-2}\right)\left(x-2\right), i.e., x+y=4$ [because this line contains the chord joining the points (2, 2) and (8, -4)].

Now, the equation of the curve containing the point $\left(a^2, a-2\right)$ is given by ${\left(y+2\right)}^2=x$. 

So, the required set of points ​$\left(a^2, a-2\right)$ within the region bounded by $x+y=4$and $y^2=2x$ is the the part of the curve ${\left(y+2\right)}^2=x$ inside the shaded region, as shown in the figure below (painted red for clarity):-



Now, we need to find the starting and ending point of the red part of the curve. As we can see from the plot, the ending point of the red part is (4, 0).

$\Rightarrow$The maximum value of a can be 2  $\left[\because \left(2^2, 2-2\right)=\left(4, 0\right)\right]$

Now, we can either find the starting point, for which we need to solve the 2 equations $y^2=2x, \mathrm{and}, {\left(y+2\right)}^2=x$, or, we can see that amongst the 4 options, only option (C) has the maximum value of a as 2.

So, the correct answer is (C).

Note:- Of course if there were more than 1 option where the maximum value of a was 2, we would have had to solve the 2 equations $y^2=2x, \mathrm{and}, {\left(y+2\right)}^2=x$ to determine the minimum value of a.