Let P be the parabola, whose focus is (-2, 1) and directrix is $$2x + y + 2 = 0$$. Then the sum of the ordinates of the points on P, whose abscissa is -2, is

The definition of a parabola says that for every point $$P(x,y)$$ on it,

distance $$PF$$ from the focus $$F(-2,1)$$ equals perpendicular distance $$PD$$ to the directrix $$2x + y + 2 = 0$$.

Write both distances using the distance formula.

$$PF = \sqrt{(x+2)^2 + (y-1)^2}$$

The perpendicular distance of $$P(x,y)$$ from the line $$2x + y + 2 = 0$$ is
$$PD = \frac{\lvert 2x + y + 2 \rvert}{\sqrt{2^2 + 1^2}} = \frac{\lvert 2x + y + 2 \rvert}{\sqrt{5}}$$

Equating and squaring (to remove the absolute value):
$$ (x+2)^2 + (y-1)^2 = \frac{(2x + y + 2)^2}{5} $$ $$-(1)$$

The required points have abscissa $$x = -2$$. Substitute $$x = -2$$ in $$(1)$$.

Left side: $$(x+2)^2 = 0$$, so
$$ (y-1)^2 $$

Right side: $$2x + y + 2 = 2(-2) + y + 2 = y - 2$$, hence
$$ \frac{(y-2)^2}{5} $$

Therefore
$$ (y-1)^2 = \frac{(y-2)^2}{5} $$ $$-(2)$$

Cross-multiply:
$$ 5(y-1)^2 = (y-2)^2 $$

Expand both squares:
$$ 5(y^2 - 2y + 1) = y^2 - 4y + 4 $$

Simplify:
$$ 5y^2 - 10y + 5 = y^2 - 4y + 4 $$
$$ 4y^2 - 6y + 1 = 0 $$ $$-(3)$$

Solve the quadratic $$4y^2 - 6y + 1 = 0$$ using the quadratic formula:

$$ y = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 4 \cdot 1}}{2 \cdot 4} = \frac{6 \pm \sqrt{36 - 16}}{8} = \frac{6 \pm 2\sqrt{5}}{8} $$

Thus the two ordinates are
$$ y_1 = \frac{3 + \sqrt{5}}{4}, \quad y_2 = \frac{3 - \sqrt{5}}{4} $$

Their sum is
$$ y_1 + y_2 = \frac{3 + \sqrt{5}}{4} + \frac{3 - \sqrt{5}}{4} = \frac{6}{4} = \frac{3}{2} $$

Hence, the sum of the ordinates of the required points is $$\frac{3}{2}$$.

Therefore, Option A is correct.