A line passing through the point $$P(\sqrt{5}, \sqrt{5})$$ intersects the ellipse $$\frac{x^2}{36} + \frac{y^2}{25} = 1$$ at A and B such that $$(PA) \cdot (PB)$$ is maximum. Then $$5(PA^2 + PB^2)$$ is equal to :

The ellipse is $$\frac{x^{2}}{36} + \frac{y^{2}}{25} = 1$$ and the external point is $$P(\sqrt{5},\sqrt{5})$$.
Let a variable line through P have slope $$m$$. Its equation can be written in parametric form as

$$x = \sqrt{5} + t ,\; y = \sqrt{5} + mt \qquad (t\in\mathbb{R})$$
The point $$P$$ corresponds to $$t = 0$$. The intersections with the ellipse (points $$A,\,B$$) occur for those values of $$t$$ that satisfy

$$\frac{(\sqrt{5}+t)^{2}}{36} + \frac{(\sqrt{5}+mt)^{2}}{25} = 1$$

Expand each square:
$$(\sqrt{5}+t)^{2}=5+2\sqrt{5}\,t+t^{2},\qquad (\sqrt{5}+mt)^{2}=5+2m\sqrt{5}\,t+m^{2}t^{2}$$

Substituting and clearing denominators by multiplying by $$900 \;(=36\!\cdot\!25)$$ gives

$$(25+36m^{2})t^{2}+\sqrt{5}(50+72m)\,t-595=0\tag{1}$$

Equation $$(1)$$ is quadratic in $$t$$, so its roots $$t_{1},t_{2}$$ represent the directed distances (along the x-axis direction of the line) from $$P$$ to $$A,\,B$$ respectively.

For a quadratic $$At^{2}+Bt+C=0$$ we have $$t_{1}t_{2}=C/A.$$
Here
$$A = 25+36m^{2},\quad C = -595 \;\Rightarrow\; t_{1}t_{2}= \frac{-595}{25+36m^{2}}$$

The actual (positive) distances are $$PA = |t_{1}|\sqrt{1+m^{2}},\; PB = |t_{2}|\sqrt{1+m^{2}}.$$
Hence

$$PA\cdot PB = |t_{1}t_{2}|(1+m^{2}) = \frac{595(1+m^{2})}{25+36m^{2}}\tag{2}$$

To maximise $$PA\cdot PB$$ we maximise the function
$$g(m)=\frac{1+m^{2}}{25+36m^{2}},\qquad m\in\mathbb{R}$$

Set $$t=m^{2}\ge 0.$$ Then $$g(t)=\frac{1+t}{25+36t}.$$ Differentiate:

$$\frac{dg}{dt}= \frac{(25+36t)\cdot1 - (1+t)\cdot36}{(25+36t)^{2}} = \frac{25+36t-36-36t}{(25+36t)^{2}} = \frac{-11}{(25+36t)^{2}} \lt 0$$

Since $$dg/dt$$ is negative for all $$t\ge 0$$, $$g(t)$$ decreases as $$t$$ increases. Therefore $$g(t)$$ (and hence $$PA\cdot PB$$) is maximum at $$t = 0$$, i.e. at $$m = 0$$.

Thus the required line is horizontal: $$y = \sqrt{5}.$$
Its intersections with the ellipse are obtained by substituting $$y=\sqrt{5}:$$

$$\frac{x^{2}}{36} + \frac{5}{25}=1 \;\Longrightarrow\; \frac{x^{2}}{36}=1-\frac{1}{5}=\frac{4}{5} \;\Longrightarrow\; x^{2}=\frac{144}{5}$$

Hence the points are $$A\bigl(\;+\frac{12}{\sqrt{5}},\,\sqrt{5}\bigr)$$ and $$B\bigl(\;-\frac{12}{\sqrt{5}},\,\sqrt{5}\bigr).$$

Compute the distances from $$P(\sqrt{5},\sqrt{5})$$:
$$PA = \left|\frac{12}{\sqrt{5}}-\sqrt{5}\right| = \left|\frac{12-5}{\sqrt{5}}\right| = \frac{7}{\sqrt{5}},$$
$$PB = \left|-\frac{12}{\sqrt{5}}-\sqrt{5}\right| = \left|\frac{-12-5}{\sqrt{5}}\right| = \frac{17}{\sqrt{5}}.$$

Now evaluate $$5(PA^{2}+PB^{2}):$$
$$PA^{2} = \frac{49}{5},\qquad PB^{2} = \frac{289}{5}$$
$$PA^{2}+PB^{2} = \frac{49+289}{5} = \frac{338}{5}$$
$$\therefore\; 5(PA^{2}+PB^{2}) = 5\cdot\frac{338}{5} = 338$$

Hence $$5(PA^{2}+PB^{2}) = 338,$$ which matches Option D.