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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?
Options:
are $\frac{\sqrt{13}}{3}$ and $\frac{10}{3}$ units respectively, then what is the length
of the transverse axis?
Solution:
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Verified Answer
The correct answer is:
$\frac{15}{2}$ unit
Length of latus rectum of a hyperbola is $\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}$ where
a is the half of the distance bet ween two vertex of the hyperbola.
Latus rectum $=\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}=\frac{10}{3}$
or, $\mathrm{b}^{2}=\frac{5 \mathrm{a}}{3}$...(1)
In case of hyperbola, $b^{2}=a^{2}\left(e^{2}-1\right)$...(2)
Putting value of $b^{2}$ from equation (1) and e $=\frac{\sqrt{13}}{3}$ in
equation (2),
$\frac{5 a}{3}=a^{2}\left(\frac{13}{9}-1\right)$
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, $a=\frac{15}{4}$
Length of transverse axis $=2 a=2 \times \frac{15}{4}=\frac{15}{2}$
a is the half of the distance bet ween two vertex of the hyperbola.
Latus rectum $=\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}=\frac{10}{3}$
or, $\mathrm{b}^{2}=\frac{5 \mathrm{a}}{3}$...(1)
In case of hyperbola, $b^{2}=a^{2}\left(e^{2}-1\right)$...(2)
Putting value of $b^{2}$ from equation (1) and e $=\frac{\sqrt{13}}{3}$ in
equation (2),
$\frac{5 a}{3}=a^{2}\left(\frac{13}{9}-1\right)$
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, $a=\frac{15}{4}$
Length of transverse axis $=2 a=2 \times \frac{15}{4}=\frac{15}{2}$
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