Let $$PQ$$ be a focal chord of the parabola $$y^2 = 36x$$ of length 100, making an acute angle with the positive $$x-$$axis. Let the ordinate of P be positive and M be the point on the line segment PQ such that $$PM : MQ = 3 : 1$$. Then which of the following points does NOT lie on the line passing through M and perpendicular to the line $$PQ$$?

The parabola is $$y^2 = 36x$$, so $$4a = 36$$ and $$a = 9$$. The focus is at $$(9, 0)$$.

Finding points P and Q:

Using the parametric form: $$P = (9t^2, 18t)$$ and $$Q = (9/t^2, -18/t)$$ where $$t > 0$$ (since the ordinate of P is positive) and $$t_1 t_2 = -1$$ for a focal chord.

The length of the focal chord is:

$$|PQ| = 9\left(t + \frac{1}{t}\right)^2 = 100$$

Let $$u = t + 1/t$$. Then $$9u^2 = 100$$, so $$u = 10/3$$.

Also $$(t - 1/t)^2 = u^2 - 4 = 100/9 - 4 = 64/9$$, so $$t - 1/t = 8/3$$ (positive for acute angle with positive $$x$$-axis).

Solving: $$t = 3$$, $$1/t = 1/3$$.

So $$P = (81, 54)$$ and $$Q = (1, -6)$$.

Verification: $$|PQ|^2 = (81-1)^2 + (54+6)^2 = 6400 + 3600 = 10000$$, so $$|PQ| = 100$$.

Finding point M:

$$PM : MQ = 3 : 1$$, so M divides PQ in ratio 3:1 from P:

$$M = \frac{1 \cdot P + 3 \cdot Q}{4} = \left(\frac{81 + 3}{4}, \frac{54 - 18}{4}\right) = (21, 9)$$

Line through M perpendicular to PQ:

Slope of PQ: $$\frac{54 - (-6)}{81 - 1} = \frac{60}{80} = \frac{3}{4}$$

Perpendicular slope: $$-\frac{4}{3}$$

Equation: $$y - 9 = -\frac{4}{3}(x - 21)$$, which simplifies to $$y = 37 - \frac{4}{3}x$$.

Checking each option:

Option A: $$(-6, 45)$$: $$y = 37 - \frac{4}{3}(-6) = 37 + 8 = 45$$. Lies on the line.

Option B: $$(6, 29)$$: $$y = 37 - \frac{4}{3}(6) = 37 - 8 = 29$$. Lies on the line.

Option C: $$(3, 33)$$: $$y = 37 - \frac{4}{3}(3) = 37 - 4 = 33$$. Lies on the line.

Option D: $$(-3, 43)$$: $$y = 37 - \frac{4}{3}(-3) = 37 + 4 = 41 \neq 43$$. Does NOT lie on the line.

The point that does NOT lie on the line is Option D: $$(-3, 43)$$.