MCQ
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 :
- A18
- B22
- C26
- D20
✓
Answer
B.
If $m \left(\sqrt{2}, \frac{4}{3}\right)$ than equation of of AB is
$T = S _1$
$\frac{x\sqrt{2}}{9}+\frac{y}{4}\left(\frac{4}{3}\right)=\frac{(\sqrt{2})^2}{9}+\frac{\left(\frac{4}{3}\right)^2}{4}$
$\frac{\sqrt{2} x}{9}+\frac{y}{3}=\frac{2}{9}+\frac{4}{9}$
$\sqrt{2} x+3 y=6 \Rightarrow y=\frac{6-\sqrt{2} x}{3}$ put in ellipse
So, $\frac{x^2}{9}+\frac{(6-\sqrt{2} x)^2}{9 \times 4}=1$
$4 x^2+36+2 x^2-12 \sqrt{2} x=36$
$6 x^2-12 \sqrt{2} x=0$
$6 x(x-2 \sqrt{2})=0$
$x=0 \& x=2 \sqrt{2}$
So $y=2 \quad y=\frac{2}{3}$
Length of chord $=\sqrt{(2 \sqrt{2}0)^2+\left(\frac{2}{3}2\right)^2}$
$=\sqrt{8+\frac{16}{9}}$
$=\sqrt{\frac{88}{9}}=\frac{2}{3} \sqrt{22}$ so $\alpha=22$
If $m \left(\sqrt{2}, \frac{4}{3}\right)$ than equation of of AB is
$T = S _1$
$\frac{x\sqrt{2}}{9}+\frac{y}{4}\left(\frac{4}{3}\right)=\frac{(\sqrt{2})^2}{9}+\frac{\left(\frac{4}{3}\right)^2}{4}$
$\frac{\sqrt{2} x}{9}+\frac{y}{3}=\frac{2}{9}+\frac{4}{9}$
$\sqrt{2} x+3 y=6 \Rightarrow y=\frac{6-\sqrt{2} x}{3}$ put in ellipse
So, $\frac{x^2}{9}+\frac{(6-\sqrt{2} x)^2}{9 \times 4}=1$
$4 x^2+36+2 x^2-12 \sqrt{2} x=36$
$6 x^2-12 \sqrt{2} x=0$
$6 x(x-2 \sqrt{2})=0$
$x=0 \& x=2 \sqrt{2}$
So $y=2 \quad y=\frac{2}{3}$
Length of chord $=\sqrt{(2 \sqrt{2}0)^2+\left(\frac{2}{3}2\right)^2}$
$=\sqrt{8+\frac{16}{9}}$
$=\sqrt{\frac{88}{9}}=\frac{2}{3} \sqrt{22}$ so $\alpha=22$
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