Mathematics - Ellipse Question with Solution | TestHub

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