Let $$a > 0$$, $$b > 0$$. Let $$e$$ and $$l$$ respectively be the eccentricity and length of the latus rectum of the hyperbola $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$. Let $$e'$$ and $$l'$$ respectively the eccentricity and length of the latus rectum of its conjugate hyperbola. If $$e^2 = \frac{11}{14}l$$ and $$(e')^2 = \frac{11}{8}l'$$, then the value of $$77a + 44b$$ is equal to

For the hyperbola $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$:

Eccentricity: $$e^2 = 1 + \frac{b^2}{a^2}$$, Latus rectum: $$l = \frac{2b^2}{a}$$

For its conjugate hyperbola $$\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$$:

Eccentricity: $$(e')^2 = 1 + \frac{a^2}{b^2}$$, Latus rectum: $$l' = \frac{2a^2}{b}$$

Set up equations from the given conditions:

From $$e^2 = \frac{11}{14}l$$:

$$1 + \frac{b^2}{a^2} = \frac{11}{14} \cdot \frac{2b^2}{a} = \frac{11b^2}{7a}$$

$$\frac{a^2 + b^2}{a^2} = \frac{11b^2}{7a}$$

$$7a(a^2 + b^2) = 11a^2 b^2 \implies 7(a^2 + b^2) = 11ab^2 \quad \cdots (1)$$

From $$(e')^2 = \frac{11}{8}l'$$:

$$1 + \frac{a^2}{b^2} = \frac{11}{8} \cdot \frac{2a^2}{b} = \frac{11a^2}{4b}$$

$$\frac{a^2 + b^2}{b^2} = \frac{11a^2}{4b}$$

$$4b(a^2 + b^2) = 11a^2 b^2 \implies 4(a^2 + b^2) = 11a^2 b \quad \cdots (2)$$

Find the ratio $$b/a$$:

Dividing equation (1) by equation (2):

$$\frac{7}{4} = \frac{11ab^2}{11a^2 b} = \frac{b}{a}$$

Therefore $$b = \frac{7a}{4}$$.

Solve for $$a$$:

Substituting $$b = \frac{7a}{4}$$ into equation (2):

$$4\left(a^2 + \frac{49a^2}{16}\right) = 11a^2 \cdot \frac{7a}{4}$$

$$4 \cdot \frac{16a^2 + 49a^2}{16} = \frac{77a^3}{4}$$

$$\frac{65a^2}{4} = \frac{77a^3}{4}$$

$$65 = 77a \implies a = \frac{65}{77} = \frac{5}{7} \cdot \frac{13}{11}$$

Find $$b$$ and compute $$77a + 44b$$:

$$b = \frac{7}{4} \cdot \frac{65}{77} = \frac{7 \times 65}{4 \times 77} = \frac{65}{44}$$

$$77a + 44b = 77 \times \frac{65}{77} + 44 \times \frac{65}{44} = 65 + 65 = 130$$

The correct answer is Option D: $$130$$.