The length of the chord of the ellipse $\frac{x^2}4+\frac{y^2}2=1$, whose mid-point is $\left(1,\frac12\right)$, is :

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] The chord with a given midpoint is $T=S_1$: $\frac x4+\frac y4=\frac14+\frac18$, i.e. $x+y=\frac32$. Substituting in the ellipse: $6x^2-12x+1=0$, so $|x_1-x_2|=\frac{\sqrt{144-24}}6=\frac{\sqrt{120}}6$. Length $=\sqrt2\cdot\frac{\sqrt{120}}6=\frac{2\sqrt{15}}3$. Answer (3).

Q. The length of the chord of the ellipse $\frac{x^2}4+\frac{y^2}2=1$, whose mid-point is $\left(1,\frac12\right)$, is :

Explanation

Reviewed by Dr. Abubacker

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