Let the sum of the focal distances of the point $$P(4, 3)$$ on the hyperbola H : $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ be $$8\sqrt{\frac{5}{3}}$$. If for $$H$$, the length of the latus rectum is $$l$$ and the product of the focal distances of the point P is m, then $$9l^2 + 6m$$ is equal to :

The hyperbola is $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$$, with foci $$F_{1}(-c,0)$$ and $$F_{2}(c,0)$$, where $$c^{2}=a^{2}+b^{2}$$.

For any point on this hyperbola, the difference of its focal distances equals the length of the transverse axis: $$|PF_{2}-PF_{1}|=2a$$.

Let
$$d_{1}=PF_{1},\qquad d_{2}=PF_{2}$$
for the point $$P(4,3)$$. The data give the following:

1. Sum of focal distances
$$d_{1}+d_{2}=8\sqrt{\frac{5}{3}}$$ $$-(1)$$

2. Using coordinates,
$$d_{1}^{2}=(4-c)^{2}+3^{2}=25-8c+c^{2}$$
$$d_{2}^{2}=(4+c)^{2}+3^{2}=25+8c+c^{2}$$

Hence
$$d_{2}^{2}-d_{1}^{2}=16c$$ $$-(2)$$

3. From the property of a hyperbola,
$$d_{2}-d_{1}=2a$$ $$-(3)$$

Using $$(1)$$ and $$(3)$$,
$$d_{2}=\frac{(d_{1}+d_{2})+(d_{2}-d_{1})}{2}=\frac{S}{2}+a,\quad d_{1}=\frac{S}{2}-a,$$ where $$S=d_{1}+d_{2}=8\sqrt{\frac{5}{3}}$$.

The product $$(d_{2}+d_{1})(d_{2}-d_{1})$$ equals $$16c$$ by $$(2)$$, so $$(d_{2}+d_{1})(d_{2}-d_{1})=S\,(2a)=16c.$$ Thus $$c=\frac{aS}{8}.$$

Squaring and substituting $$S^{2}=\left(8\sqrt{\frac{5}{3}}\right)^{2}=\frac{320}{3},$$ $$c^{2}=\frac{a^{2}S^{2}}{64}=a^{2}\,\frac{5}{3}.$$ But $$c^{2}=a^{2}+b^{2},$$ therefore $$a^{2}+b^{2}=a^{2}\,\frac{5}{3}\;\;\Longrightarrow\;\;b^{2}=\frac{2}{3}a^{2}.$$ $$-(4)$$

The point $$P(4,3)$$ lies on the hyperbola, so $$\frac{4^{2}}{a^{2}}-\frac{3^{2}}{b^{2}}=1.$$ Using $$(4)$$, $$\frac{16}{a^{2}}-\frac{9}{(2/3)a^{2}}=1 \;\;\Longrightarrow\;\; \frac{16}{a^{2}}-\frac{27}{2a^{2}}=1 \;\;\Longrightarrow\;\; \frac{2.5}{a^{2}}=1.$$ Hence $$a^{2}=\frac{5}{2},\qquad b^{2}=\frac{2}{3}\cdot\frac{5}{2}=\frac{5}{3},\qquad c^{2}=a^{2}+b^{2}=\frac{25}{6}.$$

Length of the latus rectum
For a hyperbola, $$l=\frac{2b^{2}}{a}.$$ Thus $$l=\frac{2(\frac{5}{3})}{\sqrt{\frac{5}{2}}}=\frac{10}{3}\sqrt{\frac{2}{5}},\qquad l^{2}=\frac{40}{9}.$$ Therefore $$9l^{2}=9\cdot\frac{40}{9}=40.$$

Product of the focal distances
Using $$S^{2}=d_{1}^{2}+d_{2}^{2}+2d_{1}d_{2},$$ first note $$d_{1}^{2}+d_{2}^{2}=(25-8c+c^{2})+(25+8c+c^{2})=50+2c^{2}.$$ Hence $$S^{2}=50+2c^{2}+2m \;\;\Longrightarrow\;\; m=\frac{S^{2}-50-2c^{2}}{2}.$$ Substituting $$S^{2}=\frac{320}{3},\;c^{2}=\frac{25}{6},$$ $$m=\frac{\frac{320}{3}-50-\frac{25}{3}}{2} =\frac{\frac{320-150-25}{3}}{2} =\frac{\frac{145}{3}}{2} =\frac{145}{6}.$$ Therefore $$6m=145.$$

Required value
$$9l^{2}+6m=40+145=185.$$

Hence the correct option is Option C (185).