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Let $a>0, b>0$. Let $e$ and $\ell$ 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 $e ^{\prime}$ and $\ell^{\prime}$ respectively the eccentricity and length of the latus rectum of its conjugate hyperbola. If $e ^{2}=\frac{11}{14} \ell$ and $\left( e ^{\prime}\right)^{2}=\frac{11}{8} \ell^{\prime}$, then the value of $77 a+44 b$ is equal to
$100$
$110$
$120$
$130$
Solution
$e=\sqrt{1+\frac{b^{2}}{a^{2}}}, \ell=\frac{2 b^{2}}{a}$
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 $11 \times 8 b =11 \times 8 \times \frac{4 \times 65}{16 \times 11}=130$
