Question ID: #559
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 mid point is $(\frac{5}{2},\frac{1}{2})$, then $\alpha+\beta$ is equal to:
- (1) 37
- (2) 46
- (3) 58
- (4) 72
Solution:
The equation of a chord of the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ with a given midpoint $(x_1, y_1)$ is given by $T = S_1$.
$$ \frac{xx_1}{9} + \frac{yy_1}{4} – 1 = \frac{x_1^2}{9} + \frac{y_1^2}{4} – 1 $$
Given the midpoint $(x_1, y_1) = (\frac{5}{2}, \frac{1}{2})$, we substitute these values:
$$ \frac{x(\frac{5}{2})}{9} + \frac{y(\frac{1}{2})}{4} = \frac{(\frac{5}{2})^2}{9} + \frac{(\frac{1}{2})^2}{4} $$
$$ \frac{5x}{18} + \frac{y}{8} = \frac{25}{36} + \frac{1}{16} $$
To simplify, find the LCM of the denominators (18, 8, 36, 16), which is 144. Multiply the entire equation by 144:
$$ 144 \left( \frac{5x}{18} \right) + 144 \left( \frac{y}{8} \right) = 144 \left( \frac{25}{36} \right) + 144 \left( \frac{1}{16} \right) $$
$$ 8(5x) + 18(y) = 4(25) + 9(1) $$
$$ 40x + 18y = 100 + 9 $$
$$ 40x + 18y = 109 $$
Comparing this with the given equation $\alpha x + \beta y = 109$, we get:
$$ \alpha = 40, \quad \beta = 18 $$
The value of $\alpha + \beta$ is:
$$ 40 + 18 = 58 $$
Ans. (3)
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