Let a0, b0. Let e and ell respectively be t | vedclass

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(D) For the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$,we have $e=\sqrt{1+\frac{b^{2}}{a^{2}}}$ and $\ell=\frac{2b^{2}}{a}$.Given $e^{2}=\f

Vedclass · Accepted Answer

$130$

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(D) For the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$,we have $e=\sqrt{1+\frac{b^{2}}{a^{2}}}$ and $\ell=\frac{2b^{2}}{a}$.Given $e^{2}=\frac{11}{14}\ell$,we get $1+\frac{b^{2}}{a^{2}}=\frac{11}{14} \cdot \frac{2b^{2}}{a}$,which simplifies to $\frac{a^{2}+b^{2}}{a^{2}}=\frac{11b^{2}}{7a} \dots (1)$.For the conjugate hyperbola $\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1$,we have $e^{\prime}=\sqrt{1+\frac{a^{2}}{b^{2}}}$ and $\ell^{\prime}=\frac{2a^{2}}{b}$.Given $(e^{\prime})^{2}=\frac{11}{8}\ell^{\prime}$,we get $1+\frac{a^{2}}{b^{2}}=\frac{11}{8} \cdot \frac{2a^{2}}{b}$,which simplifies to $\frac{a^{2}+b^{2}}{b^{2}}=\frac{11a^{2}}{4b} \dots (2)$.Dividing $(1)$ by $(2)$,we get $\frac{b^{2}}{a^{2}}=\frac{11b^{2}}{7a} \cdot \frac{4b}{11a^{2}} = \frac{4b^{3}}{7a^{3}}$.Thus,$7a^{3}=4b^{3} \cdot \frac{a^{2}}{b^{2}} \implies 7a=4b \dots (3)$.Substituting $a=\frac{4b}{7}$ into $(2)$: $\frac{(\frac{16b^{2}}{49})+b^{2}}{b^{2}} = \frac{11}{4} \cdot \frac{16b^{2}}{49b} \implies \frac{65}{49} = \frac{11 \cdot 4b}{49} \implies 65 = 44b \implies b = \frac{65}{44}$.Then $a = \frac{4}{7} \cdot \frac{65}{44} = \frac{65}{77}$.Finally,$77a+44b = 77(\frac{65}{77}) + 44(\frac{65}{44}) = 65+65 = 130$.

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