The length of the chord of the ellipse frac{x | vedclass

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(A) The equation of the chord of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with mid-point $(x_1, y_1)$ is given by $T=S_1$,where $T = \frac{xx_1

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$\frac{\sqrt{1691}}{5}$

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(A) The equation of the chord of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with mid-point $(x_1, y_1)$ is given by $T=S_1$,where $T = \frac{xx_1}{a^2}+\frac{yy_1}{b^2}$ and $S_1 = \frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}-1$.Given $a^2=25, b^2=16$ and $(x_1, y_1) = (1, \frac{2}{5})$.$T = \frac{x(1)}{25}+\frac{y(2/5)}{16} = \frac{x}{25}+\frac{y}{40}$.$S_1 = \frac{1^2}{25}+\frac{(2/5)^2}{16}-1 = \frac{1}{25}+\frac{4/25}{16}-1 = \frac{1}{25}+\frac{1}{100}-1 = \frac{4+1-100}{100} = -\frac{95}{100} = -\frac{19}{20}$.So,$\frac{x}{25}+\frac{y}{40} = -\frac{19}{20}$.Multiplying by $200$,we get $8x+5y = -190$,or $y = \frac{-8x-190}{5}$.Substituting this into the ellipse equation $\frac{x^2}{25}+\frac{y^2}{16}=1$ leads to a quadratic in $x$. The length of the chord is given by $L = \frac{2b}{a^2} \sqrt{a^2x_1^2+b^2y_1^2} \sqrt{a^2+b^2-x_1^2-y_1^2}$ (or using the distance formula between intersection points).Calculating the distance,we obtain $L = \frac{\sqrt{1691}}{5}$.

The length of the chord of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$,whose mid-point is $(1, \frac{2}{5})$,is equal to:

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