For the hyperbola \( H_1 \), the length of the latus rectum is given by \( \frac{2b^2}{a} \). The length of the latus rectum for \( H_1 \) is \( 15\sqrt{2} \), so we have: \[ \frac{2b^2}{a} = 15\sqrt{2}. \] Similarly, for \( H_2 \), the length of the latus rectum is \( \frac{2B^2}{A} \), and the given length is \( 12\sqrt{5} \), so: \[ \frac{2B^2}{A} = 12\sqrt{5}. \] Now, we are given that the product of the lengths of their transverse axes is \( 100\sqrt{10} \), so: \[ 2a \times 2A = 100\sqrt{10}. \] From these equations, we solve for \( e_2 \) and then compute \( 25e_2^2 \).
Final Answer: \( 25e_2^2 = 50 \).
Let a line passing through the point $ (4,1,0) $ intersect the line $ L_1: \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} $ at the point $ A(\alpha, \beta, \gamma) $ and the line $ L_2: x - 6 = y = -z + 4 $ at the point $ B(a, b, c) $. Then $ \begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix} \text{ is equal to} $
20 mL of sodium iodide solution gave 4.74 g silver iodide when treated with excess of silver nitrate solution. The molarity of the sodium iodide solution is _____ M. (Nearest Integer value) (Given : Na = 23, I = 127, Ag = 108, N = 14, O = 16 g mol$^{-1}$)