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

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Question

$T=S_1$
$\frac{x \cdot 1}{4}+\frac{y \cdot 1 / 2}{2}=\frac{1}{4}+\frac{1}{8}$
$x+y=\frac{3}{2}$
solve with ellipse
$P_{R} =\sqrt{\left(x_2-x_1\rig

Vedclass · Accepted Answer

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

Vedclass · Answer

$T=S_1$
$\frac{x \cdot 1}{4}+\frac{y \cdot 1 / 2}{2}=\frac{1}{4}+\frac{1}{8}$
$x+y=\frac{3}{2}$
solve with ellipse
$P_{R} =\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
$ =\sqrt{2}\left|x_2-x_1\right|$
image
$ y _2=\frac{3}{2}- x _2$
$y _1=\frac{3}{2}- x _1$
$y _2- y _1= x _2- x _1$
$x ^2+2 y ^2=4$
$x ^2+2\left(\frac{3}{2}- x \right)^2=4$
$6 x ^2-12 x +1=0$
$x _1+ x _2=2$
$x _1 x _2=1 / 6$
$\left|x_2-x_1\right| =\sqrt{\left(x_2+x_1\right)^2-4 x_1 x_2}$
$ =\sqrt{4-4 / 6}$
$\operatorname{PR}= \sqrt{2} \cdot 2 \cdot \frac{\sqrt{5}}{\sqrt{2} \sqrt{3}}=\frac{2}{3} \sqrt{15}$
$ =2 \sqrt{5 / 6}$

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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