If the midpoint of a chord of the ellipse x^29+y^24=1 is (√2,4/3), and the length of the c…

Question

If the midpoint of a chord of the ellipse $\frac{{x}^{2}}{9}+\frac{{y}^{2}}{4}=1$ is $(\sqrt{2},4/3)$, and the length of the chord is $\frac{2\sqrt{\alpha }}{3}$, then $\alpha $ is :

Accepted Answer

$$\begin{aligned}
&\text { If } \mathrm{m}\left(\sqrt{2}, \frac{4}{3}\right) \text { than equation of of } \mathrm{AB} \text { is }\
&\begin{aligned}
& \mathrm{T}=\mathrm{S}_1 \
& \frac{\mathrm{x} \sqrt{2}}{9}+\frac{\mathrm{y}}{4}\left(\frac{4}{3}\right)=\frac{(\sqrt{2})^2}{9}+\frac{\left(\frac{4}{3}\right)^2}{4} \
& \frac{\sqrt{2} \mathrm{x}}{9}+\frac{\mathrm{y}}{3}=\frac{2}{9}+\frac{4}{9}
\end{aligned}
\end{aligned}$$

$$\begin{aligned}
&\sqrt{2} x+3 y=6 \Rightarrow y=\frac{6-\sqrt{2} x}{3} \text { put in ellipse }\
&\begin{aligned}
& \text { So, } \frac{x^2}{9}+\frac{(6-\sqrt{2} x)^2}{9 \times 4}=1 \
& 4 x^2+36+2 x^2-12 \sqrt{2} x=36 \
& 6 x^2-12 \sqrt{2} x=0 \
& 6 x(x-2 \sqrt{2})=0 \
& x=0 & x=2 \sqrt{2} \
& \text { So } y=2 \quad y=\frac{2}{3} \
& \text { Length of chord }=\sqrt{(2 \sqrt{2}-0)^2+\left(\frac{2}{3}-2\right)^2} \
& \quad=\sqrt{8+\frac{16}{9}} \
& \quad=\sqrt{\frac{88}{9}}=\frac{2}{3} \sqrt{22} \text { so } \alpha=22
\end{aligned}
\end{aligned}$$

Solutions · 1