The length of the chord of the ellipse (x^2)/(4)+(y^2)/(2)=1, whose mid-point is (1, (1)/(…

Question

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 :

Accepted Answer

$$\begin{aligned}
&\begin{aligned}
& \mathrm{T}=\mathrm{S}_1 \
& \frac{\mathrm{x} \cdot 1}{4}+\frac{\mathrm{y} \cdot 1 / 2}{2}=\frac{1}{4}+\frac{1}{8} \
& \mathrm{x}+\mathrm{y}=\frac{3}{2}
\end{aligned}\
\&\text { solve with ellipse }\
\&\begin{aligned}
\mathrm{PR} & =\sqrt{\left(\mathrm{x}_2-\mathrm{x}_1\right)^2+\left(\mathrm{y}_2-\mathrm{y}_1\right)^2} \
& =\sqrt{2}\left|\mathrm{x}_2-\mathrm{x}_1\right|
\end{aligned}
\end{aligned}$$

$$\begin{aligned}
& y_2=\frac{3}{2}-x_2 \
& y_1=\frac{3}{2}-x_1 \
& y_2-y_1=x_2-x_1 \
& x^2+2 y^2=4 \
& x^2+2\left(\frac{3}{2}-x\right)^2=4 \
& 6 x^2-12 x+1=0 \
& x_1+x_2=2 \
& x_1 x_2=1 / 6 \
& \left|x_2-x_1\right|=\sqrt{\left(x_2+x_1\right)^2-4 x_1 x_2} \
& \quad=\sqrt{4-4 / 6} \
& P R=\sqrt{2} \cdot 2 \cdot \frac{\sqrt{5}}{\sqrt{2} \sqrt{3}}=\frac{2}{3} \sqrt{15}
\end{aligned}$$

Solutions · 1