Latus Rectum of Hyperbola

Q.

Find the equation of the hyperbola satisfying the given condition,

Vertices $(0,\pm 3),foci(0,\pm 5)$

Q. $A$ is the vertex of the hyperbola $x^2-2y^2-2\sqrt{5}x-4\sqrt{2}y-3=0$, $B$ is one of the end points of latus rectum and $C$ is the focus of the hyperbola. If $A,B$ and $C$ lies on same side of conjugate axis, then the area of the triangle $ABC$ is

1. $2$ sq. units
2. $\sqrt{\frac{3}{2}}-1$ sq. units
3. $1-\sqrt{\frac{2}{3}}$ sq. units
4. $\sqrt{\frac{3}{2}}+1$ sq. units

Q. Let $P$ be any point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ such that the absolute difference of the distances of $P$ from the two foci is $12.$ If the eccentricity of the hyperbola is $2,$ then the length of the latus rectum is

1. $4\sqrt{3}$ unit
2. $18$ unit
3. $2\sqrt{3}$ unit
4. $36$ unit