JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Hyperbola

The foci $(1,14)$ and $(1,-12)$ lie on the vertical line $x=1$, and the center is their midpoint, $(1,1)$.

Distance between foci $=2c=14-(-12)=26$, so $c=13$.

Since the hyperbola passes through $(1,6)$, which also lies on $x=1$ (the transverse axis), this point is a vertex. Distance from center $(1,1)$ to $(1,6)$ is $a=5$.

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

For a hyperbola with vertical transverse axis, $\frac{(y-k{)}^2}{a^2}-\frac{(x-h{)}^2}{b^2}=1$, the length of the latus rectum is $\ell =\frac{2b^2}{a}$: