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In a hyperbola, if the length of transverse axis is twice that of the conjugate axis, then the distance between its directrices is..... units.
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$\frac{8 b}{\sqrt{5}}$
Given, in a hyperbola, the length of transverse axis is twice that of the conjugate axis. The equation of standard hyperbola is
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\because$ Length of transverse axis $=2 \times$ length of conjugate axis
$2 a=2 \times 2 b$
$\begin{array}{lll}\Rightarrow & a & =2 b \\ \because & b^2 & =a^2\left(e^2-1\right)\end{array}$
$\begin{array}{ll}\Rightarrow & \frac{a^2}{4}=a^2\left(e^2-1\right) \\ \Rightarrow & e^2=1+\frac{1}{4}=\frac{5}{4}\end{array}$
$\Rightarrow \quad e=\frac{\sqrt{5}}{2}$
$\therefore$ Distance between its directrix $=\frac{2 a}{e}=\frac{2(2 b)}{\sqrt{5} / 2}$
$=\frac{8 b}{\sqrt{5}}$
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\because$ Length of transverse axis $=2 \times$ length of conjugate axis
$2 a=2 \times 2 b$
$\begin{array}{lll}\Rightarrow & a & =2 b \\ \because & b^2 & =a^2\left(e^2-1\right)\end{array}$
$\begin{array}{ll}\Rightarrow & \frac{a^2}{4}=a^2\left(e^2-1\right) \\ \Rightarrow & e^2=1+\frac{1}{4}=\frac{5}{4}\end{array}$
$\Rightarrow \quad e=\frac{\sqrt{5}}{2}$
$\therefore$ Distance between its directrix $=\frac{2 a}{e}=\frac{2(2 b)}{\sqrt{5} / 2}$
$=\frac{8 b}{\sqrt{5}}$
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