Let a > 0, b > 0 . Let e and l respectively be the eccentricity and...

By using eccentricty formula we get,

$e = \sqrt{1 + \dfrac{{b}^{2}}{{a}^{2}}}, l = \dfrac{2{b}^{2}}{a}$

Given ${e}^{2} = \dfrac{11}{14}l$

So, $1 + \dfrac{{b}^{2}}{{a}^{2}} = \dfrac{11}{14} \cdot \dfrac{2{b}^{2}}{a}$

$\dfrac{{a}^{2} + {b}^{2}}{{a}^{2}} = \dfrac{11}{7} \cdot \dfrac{{b}^{2}}{a} ⋯ \left(1\right)$

Also ${e}^{'} = \sqrt{1 + \dfrac{{a}^{2}}{{b}^{2}}}, {l}^{'} = \dfrac{2{a}^{2}}{b}$

Given ${\left({e}^{'}\right)}^{2} = \dfrac{11}{8}{l}^{'}$

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

$\dfrac{{a}^{2} + {b}^{2}}{{b}^{2}} = \dfrac{11}{4} \cdot \dfrac{{a}^{2}}{b} ⋯ \left(2\right)$

Now equation \left(1\right) \div$equation \left(2\right)$ we get,

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

$∴ 7a = 4b ⋯ \left(3\right)$

From \left(2\right)$</p><p>$\dfrac{\dfrac{16{b}^{2}}{49} + {b}^{2}}{{b}^{2}} = \dfrac{11}{4} \cdot \dfrac{16{b}^{2}}{49b}$</p><p>$∴ b = \dfrac{4 \times 65}{11 \times 16} ⋯ \left(4\right)$</p><p>We have to find value of</p><p>$77a + 44b$</p><p>So,$11\left(7a + 4b\right) = 11\left(4b + 4b\right) = 11 \times 8b$</p><p>$∴$Value of$11 \times 8b = 11 \times 8 \times \dfrac{4 \times 65}{16 \times 11} = 130