Let the eccentricity e of a hyperbola satisfy 6e^2-11e+3=0.

Question

Let the eccentricity $e$ of a hyperbola satisfy $6e^2-11e+3=0$. If the foci of the hyperbola are $(3,5)$ and $(3,-4)$, then the length of its latus rectum is:

Accepted Answer

$6e^2-11e+3=0 \Rightarrow (2e-3)(3e-1)=0 \Rightarrow e=\dfrac32$ or $\dfrac13$. Since $e>1$, $e=\dfrac32$.

Foci $(3,5),(3,-4)$ lie on $x=3$ (vertical transverse axis); $2c=9\Rightarrow c=\dfrac92$, and $a=\dfrac{c}{e}=\dfrac{9/2}{3/2}=3$.

$$b^2=c^2-a^2=\frac{81}{4}-9=\frac{45}{4}.$$

$$\text{Latus rectum}=\frac{2b^2}{a}=\frac{2\times45/4}{3}=\frac{15}{2}.$$

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