Let $H: y(3y+4x)=-4$ be a hyperbola and $y=mx+c$ be its conjugate axis. Let $L$ be the length of the latus rectum, $e$ be the eccentricity, and $(x_1,y_1)$ be one vertex with $y_1>0$. Then:
Step-by-Step Solution
Key Concept: Rewrite: $3y^2+4xy+4=0$. Asymptotes: $y=0$ and $3y+4x=0$. The conjugate axis is the angle bisector of the asymptotes. Bisectors of $y=0$ and $4x+3y=0$: the conjugate axis (through center, bisecting angle with $y_1>0$ direction) has slope $m=2$ (from $y=2x$).
Asymptotes $y=0$ and $3y+4x=0$. Center $O=(0,0)$. Conjugate axis slope $m=2$. $a=2/\sqrt{5}$, $b=1/\sqrt{5}$, $L=2b^2/a=1$ ✓, $e^2=5/4$ so $4e^2=5$ ✓, vertex $y_1=2/5\Rightarrow 5y_1^2=4/5\cdot 5=4$ ✓. Answer: B, C, D.
Correct Answer: BCD