Question

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

• A. 100
• B. 110
• C. 120
• D. 130

Answer 1

Answer: (D)

$e=\sqrt{1+\frac{b^{2}}{a^{2}}}, \ell=\frac{2 b^{2}}{a}$
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 $\left(e^{\prime}\right)^{2}=\frac{11}{8} \ell^{\prime}$
$1+\frac{ a ^{2}}{ b ^{2}}=\frac{11}{8} \cdot \frac{2 a ^{2}}{ b }$
$\frac{ a ^{2}+ b ^{2}}{ b ^{2}}=\frac{11}{4} \cdot \frac{ a ^{2}}{ b }$
New (1) $\div(2)$
$\frac{ b ^{2}}{ a ^{2}}=\frac{4}{7} \cdot \frac{ b ^{3}}{ a ^{3}} $
$\therefore 7 a =4 b$
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 $11 \times 8 b =11 \times 8 \times \frac{4 \times 65}{16 \times 11}=130$