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

Given the midpoint of a chord at \( \left( \sqrt{2}, \frac{4}{3} \right) \) and a chord length of \( \frac{2\sqrt{\alpha}}{3} \). The chord's equation is determined via the midpoint formula, and its length is calculated as: \[ \sqrt{2x + 3y} = 6 \Rightarrow y = \frac{6 - \sqrt{2x}}{3} \quad {(expressed in ellipse form)} \] \[ {Consequently, } \frac{x^2}{9} + \left( \frac{6 - \sqrt{2x}}{9 \times 4} \right)^2 = 1 \] \[ 4x^2 + 36 + 2x^2 - 12 \sqrt{2x} = 36 \] \[ 6x^2 - 12 \sqrt{2x} = 0 \] \[ 6x(x - \sqrt{2}) = 0 \] \[ x = 0 \quad {or} \quad x = \sqrt{2} \] Thus, \( y = 2 \) or \( y = \frac{2}{3} \). \[ {The chord's length is calculated as:} = \sqrt{\left( 2\sqrt{2} - 0 \right)^2 + \left( \frac{2}{3} - 2 \right)^2} \] \[ = \sqrt{8 + \frac{16}{9}} = \sqrt{\frac{88}{9}} = \frac{2}{3} \sqrt{22} \] \[ \Rightarrow \alpha = 22 \] Therefore, \( \alpha = 22 \).