Question

If \(\alpha x + \beta y = 109\) is the equation of the chord of the ellipse \[ \frac{x^2}{9} + \frac{y^2}{4} = 1 \] whose midpoint is \(\left(\frac{5}{2}, \frac{1}{2}\right)\), then \(\alpha + \beta\) is equal to:

• A. 37
• B. 46
• C. 58
• D. 72

Hint:

For ellipse chord problems, using the midpoint formula simplifies the process of finding the correct equation of the chord.

Answer 1

The Correct Option is C

Step 1: Equation of the chord. The equation of the chord with midpoint \((h, k)\) is: \[ T = S_1 \] Where \( T = \frac{5x}{18} + \frac{y}{8} \quad \text{and} \quad S_1 = \frac{100 + 9}{144} = 109 \) Expanding the equation: \[ 40x + 18y = 109 \] Comparing with \(\alpha x + \beta y = 109\), we get: \[\alpha = 40, \quad \beta = 18\]

Step 2: Final Calculation. \[ \alpha + \beta = 40 + 18 = 58 \]

Answer 2

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

Step 2: Formula of chord with midpoint \((x_1, y_1)\).
The equation of a chord of the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) whose midpoint is \((x_1, y_1)\) is given by:
\[ \frac{xx_1}{a^2} + \frac{yy_1}{b^2} = \frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} \]

Step 3: Substitute the known values.
Given midpoint \((x_1, y_1) = \left( \frac{5}{2}, \frac{1}{2} \right)\), \(a^2 = 9\), \(b^2 = 4\).
Substitute into the equation:
\[ \frac{x \cdot (5/2)}{9} + \frac{y \cdot (1/2)}{4} = \frac{(5/2)^2}{9} + \frac{(1/2)^2}{4} \] Simplify:
\[ \frac{5x}{18} + \frac{y}{8} = \frac{25}{36} + \frac{1}{16} \]

Step 4: Simplify the right-hand side.
Take the LCM of 36 and 16, which is 144:
\[ \frac{25}{36} + \frac{1}{16} = \frac{100}{144} + \frac{9}{144} = \frac{109}{144} \] So, the equation becomes:
\[ \frac{5x}{18} + \frac{y}{8} = \frac{109}{144} \]

Step 5: Convert to the given form \( \alpha x + \beta y = 109 \).
Multiply through by 144 to remove denominators:
\[ 144\left( \frac{5x}{18} + \frac{y}{8} \right) = 109 \] \[ 8 \times 5x + 18 \times y = 109 \] \[ 40x + 18y = 109 \] Comparing with \( \alpha x + \beta y = 109 \), we get:
\[ \alpha = 40, \, \beta = 18. \]

Step 6: Find \( \alpha + \beta \).
\[ \alpha + \beta = 40 + 18 = 58. \]

Final Answer:
\[ \boxed{58} \]