The given midpoint of the chord is \( \left( \sqrt{2}, \frac{4}{3} \right) \), and the length of the chord is \( \frac{2\sqrt{\alpha}}{3} \).
Step 2: Derive the Equation of the ChordThe equation of the chord is given as: \[ \sqrt{2x + 3y} = 6 \quad \Rightarrow \quad y = \frac{6 - \sqrt{2x}}{3} \] Substituting this into the ellipse equation: \[ \frac{x^2}{9} + \left( \frac{6 - \sqrt{2x}}{9 \times 4} \right)^2 = 1 \] Expanding and simplifying: \[ 4x^2 + 36 + 2x^2 - 12\sqrt{2x} = 36 \] Rearranging, \[ 6x^2 - 12\sqrt{2x} = 0 \] Factorizing, \[ 6x(x - \sqrt{2}) = 0 \] Thus, \[ x = 0 \quad \text{or} \quad x = \sqrt{2} \] Corresponding \(y\)-values: \[ y = 2 \quad \text{or} \quad y = \frac{2}{3} \]
Step 3: Calculate the Length of the ChordUsing the distance formula: \[ \text{Length of chord} = \sqrt{\left( 2\sqrt{2} - 0 \right)^2 + \left( \frac{2}{3} - 2 \right)^2} \] Simplifying, \[ = \sqrt{8 + \frac{16}{9}} = \sqrt{\frac{88}{9}} = \frac{2}{3} \sqrt{22} \] From the given condition, \[ \alpha = 22 \]
Final Answer: \( \alpha = 22 \)Let one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to:
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is: