If the midpoint of a chord of the ellipse x^29+y^24=1 is (√2,43), and the length of the ch…

Question

If the midpoint of a chord of the ellipse $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$ is $\left(\sqrt2,\dfrac43\right)$, and the length of the chord is $\dfrac{2\sqrt\alpha}{3}$, then $\alpha$ is:

Accepted Answer

$T=S_1$: $\dfrac{\sqrt2x}{9}+\dfrac{y}{3}=\dfrac29+\dfrac49=\dfrac23 \Rightarrow \sqrt2x+3y=6$.

Substituting $y=2-\dfrac{\sqrt2}{3}x$ into $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$ gives $\dfrac{x^2}{6}-\dfrac{\sqrt2}{3}x=0 \Rightarrow x=0$ or $2\sqrt2$.

Points: $(0,2)$ and $\left(2\sqrt2,\dfrac23\right)$.

$$\text{Chord}=\sqrt{(2\sqrt2)^2+\left(\frac23-2\right)^2}=\sqrt{8+\frac{16}{9}}=\sqrt{\frac{88}{9}}=\frac{2\sqrt{22}}{3}.$$

So $\alpha=22$.

Solutions · 1