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

Question

Let the product of the focal distances of the point $(\sqrt{3},\frac{1}{2})$ on the ellipse $\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1,(a>b)$, be $\frac{7}{4}$. Then the absolute difference of the eccentricities of two such ellipses is

Accepted Answer

\[\begin{aligned}
& \text { Product of focal distances }=\left(\mathrm{a}+\mathrm{ex}_1\right)\left(\mathrm{a}-\mathrm{ex}_1\right) \
& =\mathrm{a}^2-\mathrm{e}^2 \mathrm{x}_1^2=\mathrm{a}^2-\mathrm{e}^2(3) \
& =\mathrm{a}^2-3 \mathrm{e}^2=\frac{7}{4} \Rightarrow \mathrm{a}^2=\frac{7}{4}+3 \mathrm{e}^2 \
& \Rightarrow 4 \mathrm{a}^2=7+12 \mathrm{e}^2 \
& &\left(\sqrt{3}, \frac{1}{2}\right) \text { lines on } \frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1 \
& \therefore \frac{3}{\mathrm{a}^2}+\frac{1}{4 \mathrm{~b}^2}=1 \
& \frac{3}{\mathrm{a}^2}+\frac{1}{4\left(\mathrm{a}^2\right)\left(1-\mathrm{e}^2\right)}=1 \
& 12\left(1-\mathrm{e}^2\right)+1=4 \mathrm{a}^2\left(1-\mathrm{e}^2\right) \
& 13-12 \mathrm{e}^2=\left(7+12 \mathrm{e}^2\right)\left(1-\mathrm{e}^2\right) \
& \Rightarrow 13-12 \mathrm{e}^2=7-7 \mathrm{e}^2+12 \mathrm{e}^2-12 \mathrm{e}^4 \
& \Rightarrow 12 \mathrm{e}^4-17 \mathrm{e}^2+6=0 \
& \therefore \mathrm{e}^2=\frac{17 \pm \sqrt{289-288}}{24}=\frac{17 \pm 1}{24}=\frac{3}{4} & \frac{2}{3} \
& \therefore \mathrm{e}=\frac{\sqrt{3}}{2} & \sqrt{\frac{2}{3}} \
& \therefore \text { difference }==\frac{\sqrt{3}}{2}-\sqrt{\frac{2}{3}}=\frac{3-2 \sqrt{2}}{2 \sqrt{3}}
\end{aligned}\]

Solutions · 1