Let the product of the focal distances of the point (√3, 1 2 ) on the ellipse x^2 a^2+ y^2 b^2=1,( agt b), be 7 4. Then the absolute difference of…

$\begin{aligned} & \text { Product of focal distances }=\left(a+e x_1\right)\left(a-e x_1\right) \\ & =a^2-e^2 x_1^2=a^2-e^2(3) \\ & =a^2-3 e^2=\frac{7}{4} \Rightarrow a^2=\frac{7}{4}+3 e^2 \\ & \Rightarrow 4 a^2=7+12 e^2 \\ & \&\left(\sqrt{3}, \frac{1}{2}\right) \text { lines on } \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \\ & \therefore \frac{3}{a^2}+\frac{1}{4 b^2}=1\end{aligned}$​ Product of focal distances =(a+ex1​)(a−ex1​)=a2−e2x12​=a2−e2(3)=a2−3e2=47​⇒a2=47​+3e2⇒4a2=7+12e2&(3​,21​) lines on a2x2​+b2y2​=1∴a23​+4b21​=1​ $\begin{aligned} & \frac{3}{\mathrm{a}^2}+\frac{1}{4\left(\mathrm{a}^2\right)\left(1-\mathrm{e}^2\right)}=1 \\ & 12\left(1-\mathrm{e}^2\right)+1=4 \mathrm{a}^2\left(1-\mathrm{e}^2\right) \\ & 13-12 \mathrm{e}^2=\left(7+12 \mathrm{e}^2\right)\left(1-\mathrm{e}^2\right) \\ & \Rightarrow 13-12 \mathrm{e}^2=7-7 \mathrm{e}^2+12 \mathrm{e}^2-12 \mathrm{e}^4 \\ & \Rightarrow 12 \mathrm{e}^4-17 \mathrm{e}^2+6=0 \\ & \therefore \mathrm{e}^2=\frac{17 \pm \sqrt{289-288}}{24}=\frac{17 \pm 1}{24}=\frac{3}{4} \& \frac{2}{3} \\ & \therefore \mathrm{e}=\frac{\sqrt{3}}{2} \& \sqrt{\frac{2}{3}} \\ & \therefore \text { difference }==\frac{\sqrt{3}}{2}-\sqrt{\frac{2}{3}}=\frac{3-2 \sqrt{2}}{2 \sqrt{3}}\end{aligned}$​a23​+4(a2)(1−e2)1​=112(1−e2)+1=4a2(1−e2)13−12e2=(7+12e2)(1−e2)⇒13−12e2=7−7e2+12e2−12e4⇒12e4−17e2+6=0∴e2=2417±289−288​​=2417±1​=43​&32​∴e=23​​&32​​∴ difference ==23​​−32​​=23​3−22​​​