Question ID: #387
The equation of the chord, of the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$ whose mid-point is $(3,1)$ is:
- (1) $48x+25y=169$
- (2) $4x+122y=134$
- (3) $25x+101y=176$
- (4) $5x+16y=31$
Solution:
Equation of chord with given middle point $(x_1, y_1)$ is $T=S_{1}$.
$$ \frac{xx_1}{25} + \frac{yy_1}{16} – 1 = \frac{x_1^2}{25} + \frac{y_1^2}{16} – 1 $$
Given the mid-point $(x_1, y_1) = (3, 1)$:
$$ \frac{3x}{25} + \frac{y}{16} – 1 = \frac{3^2}{25} + \frac{1^2}{16} – 1 $$
$$ \frac{3x}{25} + \frac{y}{16} = \frac{9}{25} + \frac{1}{16} $$
Multiplying both sides by 400:
$$ 16(3x) + 25(y) = 16(9) + 25(1) $$
$$ 48x + 25y = 144 + 25 $$
$$ 48x + 25y = 169 $$
Ans. (1)
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