Question

Let $H: y(3 y+4 x)=-4$ is a hyperbola and $y=m x+c$ is its conjugate axis. Length of latus rectum of $H$ is $L$, eccentricity $e$ and $\left(x_{1}, y_{1}\right)$ is one vertex with $y_{1}>0$, then $4 e^{2}$ is equal to

Answer 1

$y=0,3 y+4 x=0$ are asymptotes whose angle

bisector are
$y=2 x, x+2 y=0$
solving $H$ with $x+2 y=0$
$y^{2}=\frac{4}{5} \Rightarrow y=\pm \frac{2}{\sqrt{5}}$
so vertex $\left(-\frac{4}{\sqrt{5}}, \frac{2}{\sqrt{5}}\right)\left(\frac{4}{\sqrt{5}},-\frac{2}{\sqrt{5}}\right)$
$C.A : y=2 x . a=2, b=1$
$OR H: \frac{\left(\frac{2 x-y}{\sqrt{5}}\right)^{2}}{4}-\frac{\left(\frac{x+2 y}{\sqrt{5}}\right)^{2}}{1}=1$
$L=1, e^{2}=\frac{5}{4}$