Question

If the midpoint of a chord of the ellipse \( \frac{x^2}{9} + \frac{y^2}{4} = 1 \) is \( \left( \sqrt{2}, \frac{4}{3} \right) \), and the length of the chord is \( \frac{2\sqrt{\alpha}}{3} \), then \( \alpha \) is:

• A. 18
• B. 22
• C. 26
• D. 20

Hint:

For ellipses, the length of a chord and the midpoint can be used together to derive key properties of the ellipse.

Answer 1

The Correct Option is B

Step 1: Identify the Given Information

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 Chord

The 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 Chord

Using 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: \( \alpha = 22 \)

Answer 2

Step 1: Write the given ellipse equation.
The equation of the ellipse is: \[ \frac{x^2}{9} + \frac{y^2}{4} = 1 \] Here, \( a^2 = 9 \) and \( b^2 = 4 \), so \( a = 3 \) and \( b = 2 \).

Step 2: Equation of chord with given midpoint.
For an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), if \( (x_1, y_1) \) is the midpoint of a chord, then the chord’s equation is: \[ \frac{x x_1}{a^2} + \frac{y y_1}{b^2} = \frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} \] Substitute \( (x_1, y_1) = \left( \sqrt{2}, \frac{4}{3} \right) \), \( a^2 = 9 \), and \( b^2 = 4 \):
\[ \frac{x(\sqrt{2})}{9} + \frac{y(\frac{4}{3})}{4} = \frac{(\sqrt{2})^2}{9} + \frac{(\frac{4}{3})^2}{4} \] Simplify: \[ \frac{\sqrt{2}x}{9} + \frac{y}{3} = \frac{2}{9} + \frac{16}{36} = \frac{2}{9} + \frac{4}{9} = \frac{6}{9} = \frac{2}{3} \] Thus, the chord equation is: \[ \frac{\sqrt{2}x}{9} + \frac{y}{3} = \frac{2}{3} \] Multiply both sides by 9: \[ \sqrt{2}x + 3y = 6 \]

Step 3: Find intersection points of chord with the ellipse.
Substitute \( y = \frac{6 - \sqrt{2}x}{3} \) into the ellipse equation:
\[ \frac{x^2}{9} + \frac{1}{4}\left(\frac{6 - \sqrt{2}x}{3}\right)^2 = 1 \] Simplify: \[ \frac{x^2}{9} + \frac{1}{36}(6 - \sqrt{2}x)^2 = 1 \] \[ \frac{x^2}{9} + \frac{1}{36}(36 - 12\sqrt{2}x + 2x^2) = 1 \] Multiply through by 36: \[ 4x^2 + 36 - 12\sqrt{2}x + 2x^2 = 36 \] Simplify: \[ 6x^2 - 12\sqrt{2}x = 0 \] \[ 6x(x - 2\sqrt{2}) = 0 \] So, \( x = 0 \) or \( x = 2\sqrt{2} \).

Step 4: Corresponding y-values.
For \( x = 0 \): \[ y = \frac{6 - 0}{3} = 2 \] For \( x = 2\sqrt{2} \): \[ y = \frac{6 - \sqrt{2}(2\sqrt{2})}{3} = \frac{6 - 4}{3} = \frac{2}{3} \] Hence, the endpoints of the chord are \( A(0, 2) \) and \( B(2\sqrt{2}, \frac{2}{3}) \).

Step 5: Find the chord length.
\[ AB = \sqrt{(2\sqrt{2} - 0)^2 + \left(\frac{2}{3} - 2\right)^2} \] \[ AB = \sqrt{8 + \left(-\frac{4}{3}\right)^2} = \sqrt{8 + \frac{16}{9}} = \sqrt{\frac{72 + 16}{9}} = \sqrt{\frac{88}{9}} = \frac{2\sqrt{22}}{3} \] Hence, the given chord length is \( \frac{2\sqrt{\alpha}}{3} \), so: \[ \alpha = 22 \]

Final Answer:
\[ \boxed{22} \]