Let the foci of a hyperbola be $$(1, 4)$$ and $$(1, -12).$$ If it passes through the point $$(1, 6)$$, then the length of its latus-rectum is :

The foci are $$F_1(1, 4)$$ and $$F_2(1, -12)$$. Because they share the same x-coordinate, the transverse axis is vertical.

The center $$(h,k)$$ is the midpoint of the foci: $$h = 1, k = \frac{4 + (-12)}{2} = -4$$.

The focal distance $$2c = 4 - (-12) = 16 \implies c = 8$$.

The absolute difference between the distances from any point on the hyperbola to its two foci is equal to the length of the major axis ($$2a$$):

$$|PF_1 - PF_2| = 2a$$

For the given point $$P(1, 6)$$:

• $$PF_1 = |6 - 4| = 2$$

• $$PF_2 = |6 - (-12)| = 18$$

$$2a = |2 - 18| = 16 \implies a = 8$$

$$a = \frac{16}{5}, \quad b^2 = \frac{2304}{25}$$

$$\text{Length of Latus Rectum} = \frac{2b^2}{a} = \frac{2 \cdot \frac{2304}{25}}{\frac{16}{5}} = \frac{4608}{25} \times \frac{5}{16} = \frac{288}{5}$$