Given: The triangle \( OAB \) is an equilateral triangle, and the point \( P \) divides \( AB \) in the ratio \( 2:3 \).
\( \cos 60^\circ = \frac{OA^2 + AP^2 - OP^2}{2 \cdot OA \cdot AP} \).
\( \frac{1}{2} = \frac{\lambda^2 + \frac{4\lambda^2}{25} - OP^2}{2 \cdot \lambda \cdot \frac{2\lambda}{5}} \).
\( \frac{1}{2} = \frac{\lambda^2 + \frac{4\lambda^2}{25} - OP^2}{\frac{4\lambda^2}{5}} \).
\( \frac{2\lambda^2}{5} = \lambda^2 + \frac{4\lambda^2}{25} - OP^2 \).
\( OP^2 = \lambda^2 + \frac{4\lambda^2}{25} - \frac{2\lambda^2}{5} \).
\( OP^2 = \frac{25\lambda^2}{25} + \frac{4\lambda^2}{25} - \frac{10\lambda^2}{25} \).
\( OP^2 = \frac{19\lambda^2}{25} \).
\( OP = \sqrt{\frac{19\lambda^2}{25}} = \frac{\sqrt{19}}{5} \lambda \).
Final Answer: \( OP = \frac{\sqrt{19}}{5} \lambda \).
Let circle \( C \) be the image of
\[ x^2 + y^2 - 2x + 4y - 4 = 0 \]
in the line
\[ 2x - 3y + 5 = 0 \]
and \( A \) be the point on \( C \) such that \( OA \) is parallel to the x-axis and \( A \) lies on the right-hand side of the centre \( O \) of \( C \).
If \( B(\alpha, \beta) \), with \( \beta < 4 \), lies on \( C \) such that the length of the arc \( AB \) is \( \frac{1}{6} \) of the perimeter of \( C \), then \( \beta - \sqrt{3}\alpha \) is equal to:
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