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Let $\mathrm{a}>0, \mathrm{~b}>0$ . Let $e$ and $l$ respectively be the eccentricity and length of the latus rectum of the hyperbola $\mathrm{\frac{x^{2}}{\mathrm{a}^{2}}-\frac{y^{2}}{\mathrm{~b}^{2}}=1}$ . Let $\mathrm{e}^{\prime}$ and $l^{\prime}$ respectively be the

eccentricity and length of the latus rectum of its conjugate hyperbola. If

$\mathrm{e}^{2}=\frac{11}{14} l$ and $\left(\mathrm{e}^{\prime}\right)^{2}=\frac{11}{8} l^{\prime}$ , then the value of $77 \mathrm{a}+44 \mathrm{~b}$ is equal to :
Option: 1
$100$

Option: 2
$110$

Option: 3
$120$

Option: 4
$130$

Answers (1)

best_answer

$\mathrm{\text { Given } e^{2}=\frac{11}{14} \: l} \\$

$\mathrm{\Rightarrow 1+\frac{b^{2}}{a^{2}}=\frac{11}{14} \cdot\left(\frac{2 b^{2}}{a}\right)} \\$

$\mathrm{\Rightarrow \frac{a^{2}+b^{2}}{a^{2}}=\frac{22 b^{2}}{14 a} }\\$

$\mathrm{\Rightarrow 7\left(a^{2}+b^{2}\right)=11 b^{2} a}$ ...........(i)

$\mathrm{\text { Also }e'^{2}=\frac{11}{8}\: l^{\prime}} \\$

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

$\mathrm{\Rightarrow b^{2}+a^{2}=b^{2}\left(\frac{11 a^{2}}{4 b}\right)} \\$

$\mathrm{\Rightarrow 4\left(a^{2}+b^{2}\right)=11 a^{2} b}$ .............(ii)

Divide eqn (i) and (ii)

$\mathrm{\frac{7}{4}=\frac{b}{a}} \\$

$\mathrm{\text { Also } 4\left(a^{2}+b^{2}\right)=11 a^{2} b} \\$

$\mathrm{\Rightarrow 4\left(1+\left(\frac{b}{a}\right)^{2}\right)=11 b}$

$\mathrm{\Rightarrow 11 b=4\left(1+\frac{49}{16}\right) }\\$

$\mathrm{\Rightarrow 11 b=4\left(\frac{65}{16}\right)} \\$

$\mathrm{\Rightarrow b=\frac{65}{44}} \\$

$\mathrm{\Rightarrow a=\frac{4 b}{7}=\frac{4}{7} \cdot \frac{65}{44}=\frac{65}{77} }\\$

$\mathrm{\therefore 77 a+44 b=65+65=130}$

Hence the correct answer is option 4.

Posted by

vishal kumar

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