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

Given $e ^{2}=\frac{11}{14} \ell$

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

$\frac{a^{2}+b^{2}}{a^{2}}=\frac{11}{7} \cdot \frac{b^{2}}{a}$……..$(1)$

Also $e ^{\prime}=\sqrt{1+\frac{ a ^{2}}{ b ^{2}}}, \ell^{\prime}=\frac{2 a ^{2}}{ b }$

Given $\left( e ^{\prime}\right)^{2}=\frac{11}{8} \ell^{\prime}$

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

$\frac{ a ^{2}+ b ^{2}}{ b ^{2}}=\frac{11}{4} \cdot \frac{ a ^{2}}{ b }…………(2)$

New $(1)$ $\div$ $(2)$

$\frac{ b ^{2}}{ a ^{2}}=\frac{4}{7} \cdot \frac{ b ^{3}}{ a ^{3}}$

$\therefore 7 a =4 b…….(3)$

From $(2)$

$\frac{\frac{16 b ^{2}}{49}+ b ^{2}}{ b ^{2}}=\frac{11}{4} \cdot \frac{16 b ^{2}}{49 b }$

$\frac{65}{49}=\frac{11}{4} \cdot \frac{16}{49} \cdot b$

$\therefore b =\frac{4 \times 65}{11 \times 16}……….(4)$

We have to find value of

$77 a+44 b$

$11(7 a +4 b )=11(4 b +4 b )=11 \times 8 b$

$\therefore$ Value of $11 \times 8 b =11 \times 8 \times \frac{4 \times 65}{16 \times 11}=130$