Step 1: Consider the equation of the ellipse \( \frac{x^2}{4} + \frac{y^2}{9} = 1 \).
Step 2: Knowing the midpoint of the chord \((1,1)\) is on the ellipse, the line \(x + \alpha y = \beta\) must pass through this point.
Step 3: Plugging \((1,1)\) into the line equation gives: \[ 1 + \alpha \times 1 = \beta \quad \Rightarrow \quad \beta = 1 + \alpha. \]
An inductor and a resistor are connected in series to an AC source of voltage \( 144\sin(100\pi t + \frac{\pi}{2}) \) volts. If the current in the circuit is \( 6\sin(100\pi t + \frac{\pi}{2}) \) amperes, then the resistance of the resistor is: