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: \( 25e_2^2 = 50 \).
In Carius method of estimation of halogen, 0.25 g of an organic compound gave 0.15 g of silver bromide (AgBr). The percentage of Bromine in the organic compound is ________________ × 10-1% (Nearest integer).
(Given: Molar mass of Ag is 108 and Br is 80 g mol-1)
For the reaction:
The correct order of set of reagents for the above conversion is :
Match List - I with List - II.