Let the product of the focal distances of the | vedclass

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Question

$	ext { Product of focal distances }=\left(a+e x_1ight)\left(a-e x_1ight)$
$= a ^2- e ^2 x _1^2= a ^2- e ^2(3)$
$= a ^2-3 e ^2=\frac{7}{4} \Rig

Vedclass · Accepted Answer

$\frac{3-2 \sqrt{2}}{2 \sqrt{3}}$

Vedclass · Answer

$	ext { Product of focal distances }=\left(a+e x_1ight)\left(a-e x_1ight)$
$= a ^2- e ^2 x _1^2= a ^2- e ^2(3)$
$= a ^2-3 e ^2=\frac{7}{4} \Rightarrow a ^2=\frac{7}{4}+3 e ^2$
$\Rightarrow 4 a ^2=7+12 e ^2$
$\ \left(\sqrt{3}, \frac{1}{2}ight) 	ext { lines on } \frac{ x ^2}{ a ^2}+\frac{ y ^2}{b^2}=1$
$	herefore \frac{3}{ a ^2}+\frac{1}{4 b^2}=1$
$\frac{3}{ a ^2}+\frac{1}{4\left( a ^2ight)\left(1- e ^2ight)}=1$
$12\left(1- e ^2ight)+1=4 a ^2\left(1- e ^2ight)$
$13-12 e ^2=\left(7+12 e ^2ight)\left(1- e ^2ight)$
$\Rightarrow 13-12 e ^2=7-7 e ^2+12 e ^2-12 e ^4$
$\Rightarrow 12 e ^4-17 e ^2+6=0$
$	herefore e ^2=\frac{17 \pm \sqrt{289-288}}{24}=\frac{17 \pm 1}{24}=\frac{3}{4} \ \frac{2}{3}$
$	herefore e =\frac{\sqrt{3}}{2} \ \sqrt{\frac{2}{3}}$
$	herefore 	ext { difference }==\frac{\sqrt{3}}{2}-\sqrt{\frac{2}{3}}=\frac{3-2 \sqrt{2}}{2 \sqrt{3}}$

Let the product of the focal distances of the point $\left(\sqrt{3}, \frac{1}{2}\right)$ 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

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