Question

Let $H:y(3y+4x)=-4$ is a hyperbola and $y=mx+c$ 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 $4e^2$ is equal to

Answer 1

Answer: 5

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

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