Let one focus of the hyperbola \( H : \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 \) be at \( (\sqrt{10}, 0) \) and the corresponding directrix be \( x = \dfrac{9}{\sqrt{10}} \). If \( e \) and \( l \) respectively are the eccentricity and the length of the latus rectum of \( H \), then \( 9 \left(e^2 + l \right) \) is equal to:
The problem asks for the value of the expression \( 9(e^2 + l) \) for a given hyperbola \( H: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \). We are provided with the coordinates of one focus, \( (\sqrt{10}, 0) \), and the equation of the corresponding directrix, \( x = \frac{9}{\sqrt{10}} \). Here, \( e \) represents the eccentricity and \( l \) represents the length of the latus rectum of the hyperbola.
For a standard horizontal hyperbola with the equation \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), we use the following standard properties:
Step 1: Extract equations from the given information.
We are given that one focus of the hyperbola is at \( (\sqrt{10}, 0) \). By comparing this with the standard form of a focus \( (ae, 0) \), we can write our first equation:
\[ ae = \sqrt{10} \quad \cdots(1) \]
We are also given that the corresponding directrix is the line \( x = \frac{9}{\sqrt{10}} \). Comparing this with the standard equation of a directrix \( x = \frac{a}{e} \), we get our second equation:
\[ \frac{a}{e} = \frac{9}{\sqrt{10}} \quad \cdots(2) \]
Step 2: Calculate the values of \( a \) and \( e \).
To find the value of \( a \), we can multiply equation (1) and equation (2):
\[ (ae) \times \left(\frac{a}{e}\right) = \sqrt{10} \times \frac{9}{\sqrt{10}} \] \[ a^2 = 9 \]
Therefore, the semi-transverse axis is \( a = 3 \).
Now, substitute the value of \( a = 3 \) back into equation (1) to find the eccentricity \( e \):
\[ 3e = \sqrt{10} \implies e = \frac{\sqrt{10}}{3} \]
Next, we find the value of \( e^2 \):
\[ e^2 = \left(\frac{\sqrt{10}}{3}\right)^2 = \frac{10}{9} \]
Step 3: Calculate the value of \( b^2 \).
We use the relationship \( b^2 = a^2(e^2 - 1) \) and substitute the values of \( a^2 = 9 \) and \( e^2 = \frac{10}{9} \):
\[ b^2 = 9 \left( \frac{10}{9} - 1 \right) = 9 \left( \frac{10 - 9}{9} \right) = 9 \left( \frac{1}{9} \right) \] \[ b^2 = 1 \]
Step 4: Calculate the length of the latus rectum (\( l \)).
The formula for the latus rectum is \( l = \frac{2b^2}{a} \). Substituting the values of \( b^2 = 1 \) and \( a = 3 \):
\[ l = \frac{2(1)}{3} = \frac{2}{3} \]
The final step is to compute the value of the expression \( 9(e^2 + l) \). We substitute the calculated values of \( e^2 = \frac{10}{9} \) and \( l = \frac{2}{3} \).
\[ 9(e^2 + l) = 9 \left( \frac{10}{9} + \frac{2}{3} \right) \]
To add the fractions inside the parenthesis, we find a common denominator:
\[ 9 \left( \frac{10}{9} + \frac{6}{9} \right) = 9 \left( \frac{10 + 6}{9} \right) = 9 \left( \frac{16}{9} \right) \]
The 9 in the numerator cancels out the 9 in the denominator:
\[ 9(e^2 + l) = 16 \]
Hence, the value of the expression \( 9(e^2 + l) \) is 16.
Let \( ae = \sqrt{10} \), and the directrix is at \( x = \frac{a}{e} = \frac{\sqrt{10}}{2} \Rightarrow a = \sqrt{10}, e = 2 \).
Now, \( b^2 = a^2(e^2 - 1) = 10(4 - 1) = 30 \)
So \( l = \frac{2b^2}{a} = \frac{60}{\sqrt{10}} = 6\sqrt{10} \).
But instead, based on the final calculation logic shared: \[ 9(e^2 + l) = 9\left( \frac{10}{9} \right) = 16 \]
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,