The length of the chord of the ellipse x^24+y^22=1, whose midpoint is (1,12), is:

Question

The length of the chord of the ellipse $\dfrac{x^2}{4}+\dfrac{y^2}{2}=1$, whose midpoint is $\left(1,\dfrac12\right)$, is:

Accepted Answer

$T=S_1$: $\dfrac{x}{4}+\dfrac{y}{4}=\dfrac14+\dfrac18=\dfrac38 \Rightarrow x+y=\dfrac32$.

Substituting $y=\tfrac32-x$ into $x^2+2y^2=4$: $6x^2-12x+1=0$, so $x_1+x_2=2,\ x_1x_2=\tfrac16$.

$$(x_2-x_1)^2=4-\frac46=\frac{10}{3}.$$

Chord length $=\sqrt2\,|x_2-x_1|=\sqrt2\sqrt{10/3}=\sqrt{20/3}=\dfrac{2\sqrt{15}}{3}$.

Solutions · 1