Centre of the circle:
\[ \left( \frac{3}{2}, 1 \right) \]
Equation of diameter:
\[ 2\left( \frac{3}{2} \right) + 3(1) - k = 0 \implies k = 6 \]
Now, equation of ellipse becomes:
\[ x^2 + 9y^2 = 36 \]
\[ \frac{x^2}{6^2} + \frac{y^2}{2^2} = 1 \]
Length of latus rectum (LR):
\[ LR = \frac{2b^2}{a} = \frac{2 \cdot 2^2}{6} = \frac{8}{6} = \frac{4}{3} = \frac{m}{n} \]
Thus, \[ 2m + n = 2(4) + 3 = 11 \]
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32