JEE Main 2025 Jan 22 Shift 1, Mathematics Q20: Introduction to Hyperbola

Foci lie on the vertical line $x=1$. Centre is the midpoint: $(1,1)$. Distance from centre to each focus: $c=13$ (along the y-axis).

The point $(1,6)$ lies on the transverse axis (same $x$ as foci). For a vertical hyperbola with semi-transverse axis $b$ (along y), the vertices are at $(1,1\pm b)$, so $(1,6)$ being on the hyperbola through the axis suggests it is a vertex: $b=5$.

Then $a^2=c^2-b^2=169-25=144$, $a=12$.

Length of latus rectum of a hyperbola $\frac{(y-k{)}^2}{b^2}-\frac{(x-h{)}^2}{a^2}=1$ is $\frac{2a^2}{b}=\frac{2\cdot 144}{5}=\frac{288}{5}$.