If $e_1$ and $e_2$ are the eccentricities of the hyperbola $16 x^2-9 y^2=1$ and its conjugate respectively. Then, $3 e_1=$

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Aicharya · Accepted Answer

Given, hyperbola is $16 x^2-9 y^2=1$ $\frac{x^2}{\frac{1}{16}}-\frac{y^2}{\frac{1}{9}}=1$ $\Rightarrow \quad \frac{x^2}{\left(\frac{1}{4}ight)^2}-\frac{y^2}{\left(\frac{1}{3}ight)^2}=1$ We know that Eccentricity of a hyperbola, $e_1^2=1+\frac{b^2}{a^2}$ $\begin{aligned} & \Rightarrow e_1^2=1+\frac{1 	imes 16}{9 	imes 1} \ & \Rightarrow e_1^2=\frac{25}{9}\end{aligned}$ $\Rightarrow \quad e_1=\frac{5}{3}$ ... (i) Now, eccentricity of its conjugate $e_2^2=1+\frac{a^2}{b^2}$ $\begin{array}{ll}\Rightarrow & e_2^2=1+\frac{1 	imes 9}{16 	imes 1} \ \Rightarrow & e_2^2=\frac{25}{16}\end{array}$ $\Rightarrow \quad e_2=\frac{5}{4}$ ...(ii) From Eqs. (i) and (ii), $3 e_1=3 	imes \frac{5}{3}=\frac{5}{4} 	imes 4$ $\Rightarrow \quad 3 e_1=4 e_2 \quad\left[\because e_2=\frac{5}{4}ight]$

If $e_1$ and $e_2$ are the eccentricities of the hyperbola $16 x^2-9 y^2=1$ and its conjugate respectively.…

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