We are given:
We know the formula for the latus rectum of an ellipse is:
\[ L = \frac{2b^2}{a} \]
From the problem, we are given \( L = \sqrt{14} \), so:
\[ \frac{2b^2}{a} = \sqrt{14} \]
Thus,
\[ 2b^2 = a \sqrt{14} \tag{1} \]
Next, use the relationship for the eccentricity \( e \) of an ellipse:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} = \frac{1}{\sqrt{2}} \]
Squaring both sides:
\[ e^2 = 1 - \frac{b^2}{a^2} = \frac{1}{2} \]
This implies:
\[ \frac{b^2}{a^2} = \frac{1}{2} \]
Substitute this into equation (1):
\[ 2 \left(\frac{a^2}{2}\right) = a \sqrt{14} \] \[ a^2 = a \sqrt{14} \]
Thus:
\[ a = \sqrt{14} \] 1
Substitute \( a = \sqrt{14} \) into \( b^2 = \frac{a^2}{2} \):
\[ b^2 = \frac{14}{2} = 7 \]
Now, compute the square of the eccentricity:
\[ e^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{7}{14} = \frac{7}{14} = \frac{1}{2} \]
Thus, the square of the eccentricity is \( \frac{3}{2} \).
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