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If the eccentricity and length of latus rectum of a hyperbola are $\frac{\sqrt{13}}{3}$ and $\frac{10}{3}$ units respectively, then what is the length of the transverse axis?
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$\frac{15}{4}$ unit
Length of latus rectum of a hyperbola is $\frac{2 b^2}{a}$ where $a$ is the half of the distance between two vertex of the hyperbola.
Latus rectum $=\frac{2 b^2}{a}=\frac{10}{3}$ or, $b^2=\frac{5 a}{3}$ ...(i)
In case of hyperbola, $\mathrm{b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)$ ...(ii)
Putting value of $b^2$ from equation (i) and $e=$
$\frac{\sqrt{13}}{3}$ in equation (ii), $\frac{5 \mathrm{a}}{3}=\mathrm{a}^2\left(\frac{13}{9}-1\right)$ or,
$\frac{5 a}{3}=\frac{4 a^2}{9}$
$\Rightarrow 4 a^2-15 a=0$ or, $a(4-15 a)=0$
$a \neq 0$, hence, $a=\frac{15}{4}$
Latus rectum $=\frac{2 b^2}{a}=\frac{10}{3}$ or, $b^2=\frac{5 a}{3}$ ...(i)
In case of hyperbola, $\mathrm{b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)$ ...(ii)
Putting value of $b^2$ from equation (i) and $e=$
$\frac{\sqrt{13}}{3}$ in equation (ii), $\frac{5 \mathrm{a}}{3}=\mathrm{a}^2\left(\frac{13}{9}-1\right)$ or,
$\frac{5 a}{3}=\frac{4 a^2}{9}$
$\Rightarrow 4 a^2-15 a=0$ or, $a(4-15 a)=0$
$a \neq 0$, hence, $a=\frac{15}{4}$
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