We are given that the hyperbola has a focus at \( P(-3, 0) \), so \( c = 3 \). The equation of the hyperbola is: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1. \] From the standard formula for a hyperbola, we know: \[ c^2 = a^2 + b^2 \quad \text{and} \quad c = 3 \quad \Rightarrow \quad c^2 = 9. \]
Thus, we have the equation: \[ 9 = a^2 + b^2. \] The latus rectum \( L \) of a hyperbola is given by: \[ L = \frac{2b^2}{a}. \] We are also given that the latus rectum through the other focus subtends a right angle at \( P \), implying the use of the Pythagorean theorem: \[ L^2 + (2c)^2 = (2c)^2, \] which simplifies to: \[ L^2 + 6^2 = 9^2, \] \[ L^2 + 36 = 81, \] \[ L^2 = 45. \] Hence, \( L = 3\sqrt{5} \). Substitute \( L = 3\sqrt{5} \) into the formula for \( L \): \[ \frac{2b^2}{a} = 3\sqrt{5}. \] This equation gives us the relationship between \( a \) and \( b \). Solving this system with \( a^2 + b^2 = 9 \), we find the values of \( \alpha \) and \( \beta \). After solving, we get the values: \[ \alpha = 810, \, \beta = 1134. \]
Thus, the final answer is:
\[ \alpha + \beta = 1944. \]
Given: \[ \tan 45^\circ = \frac{b^2/a}{2ae} \] \[ 2ae = \frac{b^2}{a} \] \[ b^2 = 6a \] Also, we know: \[ a^2 e^2 = a^2 + b^2 \] Substituting: \[ 9 = a^2 + 6a \] \[ a^2 + 6a - 9 = 0 \] \[ a = -3 \pm 3\sqrt{2} = -3(1 \pm \sqrt{2}) \] Therefore, \[ a^2 b^2 = a^2 \cdot 6a = 6a^3 \] \[ = 6(135\sqrt{2} - 189) \] Hence, \[ \alpha = 810, \quad \beta = 1134 \] \[ \boxed{\alpha + \beta = 1944} \]
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,