Let the product of the focal distances of the point $\left(\sqrt3,\frac12\right)$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,(ab)$, be $\frac74$. Then the absolute difference of the eccentricities of two such ellipses is

Question

Let the product of the focal distances of the point $\left(\sqrt3,\frac12\right)$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,(a>b)$, be $\frac74$. Then the absolute difference of the eccentricities of two such ellipses is

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] Product of focal distances $=(a-ex)(a+ex)=a^2-e^2x^2=a^2-\frac{3(a^2-b^2)}{a^2}=\frac74$, and the point lies on the ellipse: $\frac3{a^2}+\frac1{4b^2}=1$. Solving gives $(a^2,b^2)=(4,1)$ or $\left(\frac{15}4,\frac54\right)$. Then $e_1^2=\frac34$ and $e_2^2=1-\frac13=\frac23$, so $|e_1-e_2|=\frac{\sqrt3}2-\sqrt{\frac23}=\frac{3-2\sqrt2}{2\sqrt3}$. Answer (2).

Q. Let the product of the focal distances of the point $\left(\sqrt3,\frac12\right)$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,(a>b)$, be $\frac74$. Then the absolute difference of the eccentricities of two such ellipses is

Explanation

Reviewed by Dr. Abubacker

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