Let the product of the focal distances of the point (√3,12) on the ellipse x^2a^2+y^2b^2=1…

Question

Let the product of the focal distances of the point $\left(\sqrt3,\dfrac12\right)$ on the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\ (a>b)$ be $\dfrac74$. Then the absolute difference of the eccentricities of two such ellipses is:

Accepted Answer

Product of focal distances $=(a+ex_1)(a-ex_1)=a^2-e^2x_1^2=a^2-3e^2=\dfrac74 \Rightarrow 4a^2=7+12e^2$.

Point on ellipse: $\dfrac{3}{a^2}+\dfrac{1}{4b^2}=1$, with $b^2=a^2(1-e^2)$:

$$12(1-e^2)+1=4a^2(1-e^2) \;\Rightarrow\; 13-12e^2=(7+12e^2)(1-e^2) \;\Rightarrow\; 12e^4-17e^2+6=0.$$

$$e^2=\frac{17\pm1}{24}=\frac34 \text{ or } \frac23 \;\Rightarrow\; e=\frac{\sqrt3}{2} \text{ or } \sqrt{\frac23}.$$

$$\left|\frac{\sqrt3}{2}-\sqrt{\frac23}\right|=\frac{3-2\sqrt2}{2\sqrt3}.$$

Solutions · 1