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

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(A) The equation of the chord of the ellipse $\frac{x^2}{4}+\frac{y^2}{2}=1$ with mid-point $(h, k) = \left(1, \frac{1}{2}\right)$ is given by $T = S_

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$\frac{2}{3} \sqrt{15}$

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(A) The equation of the chord of the ellipse $\frac{x^2}{4}+\frac{y^2}{2}=1$ with mid-point $(h, k) = \left(1, \frac{1}{2}\right)$ is given by $T = S_1$.$\frac{x(1)}{4} + \frac{y(1/2)}{2} = \frac{1^2}{4} + \frac{(1/2)^2}{2}$$\frac{x}{4} + \frac{y}{4} = \frac{1}{4} + \frac{1}{8}$$\frac{x+y}{4} = \frac{3}{8} \implies x+y = \frac{3}{2} \implies y = \frac{3}{2} - x$.Substituting $y = \frac{3}{2} - x$ into the ellipse equation $\frac{x^2}{4} + \frac{y^2}{2} = 1$:$x^2 + 2y^2 = 4$$x^2 + 2\left(\frac{3}{2} - x\right)^2 = 4$$x^2 + 2\left(\frac{9}{4} - 3x + x^2\right) = 4$$x^2 + \frac{9}{2} - 6x + 2x^2 = 4$$3x^2 - 6x + \frac{1}{2} = 0 \implies 6x^2 - 12x + 1 = 0$.Let the roots be $x_1, x_2$. Then $x_1+x_2 = 2$ and $x_1x_2 = \frac{1}{6}$.$|x_2 - x_1| = \sqrt{(x_1+x_2)^2 - 4x_1x_2} = \sqrt{4 - 4(\frac{1}{6})} = \sqrt{4 - \frac{2}{3}} = \sqrt{\frac{10}{3}}$.Since $y = \frac{3}{2} - x$,we have $|y_2 - y_1| = |x_1 - x_2| = \sqrt{\frac{10}{3}}$.The length of the chord is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} = \sqrt{\frac{10}{3} + \frac{10}{3}} = \sqrt{\frac{20}{3}} = 2\sqrt{\frac{5}{3}} = \frac{2\sqrt{15}}{3}$.

The length of the chord of the ellipse $\frac{x^2}{4}+\frac{y^2}{2}=1$,whose mid-point is $\left(1, \frac{1}{2}\right)$,is:

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