1. Given hyperbola

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

For this hyperbola:

• Eccentricity: $e=\sqrt{1+\frac{b^2}{a^2}}=\frac{\sqrt{a^2+b^2}}{a}$
• Length of latus rectum: $l=\frac{2b^2}{a}$

The condition given is $e^2=\frac{11}{14}l$

Now, $e^2=1+\frac{b^2}{a^2}=\frac{a^2+b^2}{a^2}$

and $\frac{11}{14}l=\frac{11}{14}\cdot \frac{2b^2}{a}=\frac{11b^2}{7a}$

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

Multiply by $7a^2$: $7(a^2+b^2)=11ab^2$

So we get $7a^2+7b^2=11ab^2 \qquad (1)$

2. Conjugate hyperbola

The conjugate hyperbola is $\frac{y^2}{b^2}-\frac{x^2}{a^2}=1$

For this hyperbola:

• Transverse semi-axis $=b$
• Conjugate semi-axis $=a$

Hence,

• Eccentricity: $e'=\sqrt{1+\frac{a^2}{b^2}}=\frac{\sqrt{a^2+b^2}}{b}$
• Length of latus rectum: $l'=\frac{2a^2}{b}$

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

Now, $(e')^2=1+\frac{a^2}{b^2}=\frac{a^2+b^2}{b^2}$

and $\frac{11}{8}l'=\frac{11}{8}\cdot \frac{2a^2}{b}=\frac{11a^2}{4b}$

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

Multiply by $4b^2$: $4(a^2+b^2)=11a^2b$

So, $4a^2+4b^2=11a^2b \qquad (2)$

3. Solve the system

From (1): $7(a^2+b^2)=11ab^2$

From (2): $4(a^2+b^2)=11a^2b$

Let $S=a^2+b^2$

Then, $7S=11ab^2 \qquad (3)$ $4S=11a^2b \qquad (4)$

Divide (3) by (4): $\frac{7}{4}=\frac{11ab^2}{11a^2b}=\frac{b}{a}$

Hence, $b=\frac{7a}{4}$

Substitute into (4): $4(a^2+b^2)=11a^2b$

Using $b=\frac{7a}{4}$, $a^2+b^2=a^2+\frac{49a^2}{16}=\frac{65a^2}{16}$

So, $4\cdot \frac{65a^2}{16}=11a^2\cdot \frac{7a}{4}$

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

Multiply by 4: $65a^2=77a^3$

Since $a>0$, $a=\frac{65}{77}$

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

4. Compute the required value

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

$=65+65=130$

5. Final answer

The value of $77a+44b$ is

$\boxed{130}$

So the correct option is D.