Question

If the eccentricity and length of latus rectum of a hyperbola are $\frac{\sqrt{1}3}{3}$ and $\frac{10}{3}$ units respectively, then what is the length of the traverse axis?

• A. $\frac{7}{2}$ units
• B. $12$ units
• C. $\frac{15}{2}$ units
• D. $\frac{15}{4}$ units

Answer 1

The correct option is C $\frac{15}{2}$ units

$formulaused$
$L=\frac{2b^2}{a}$
$e=\sqrt{1-\frac{b^2}{a^2}}$
$Lengthoftransverseaxis=2a$
$Given$
$e=\frac{\sqrt{13}}{3},L=\frac{10}{3}$
$\frac{2b^2}{a}=\frac{10}{3}$
$\Rightarrow b^2=\frac{5a}{3}$
$b^2=a^2(e^2-1)$
$\Rightarrow \frac{5a}{3}=a^2(\frac{13}{9}-1)$
$\Rightarrow \frac{5a}{3}=a^2\left(\frac{4}{9}\right)$
$\Rightarrow 4a^2-15a=0$
$\Rightarrow a(4a-15)=0$
$a=0,\frac{15}{4}$
$a\ne 0$
$Hencea=\frac{15}{4}$
$Lengthoftransverseaxis=2a$
$=\frac{15}{4}\left(2\right)$
$=\frac{15}{2}$