Let the distance between two parallel lines be 5 units and a point $P$ lie between the lines at a unit distance from one of them. An equilateral triangle PQR is formed such that $Q$ lies on one of the parallel lines, while $R$ lies on the other. Then $(Q R)^2$ is equal to $\_\_\_\_$ .
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Solution
$$
\begin{aligned}
& \frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1 \text { foci are }(\mathrm{ae}, 0) \text { and }(-\mathrm{ae}, 0) \\
& \frac{\mathrm{x}^2}{\mathrm{~A}^2}+\frac{\mathrm{y}^2}{\mathrm{~B}^2}=1 \text { foci are }\left(\mathrm{Ae}^{\prime}, 0\right) \text { and }\left(-\mathrm{Ae}^{\prime}, 0\right) \\
& \Rightarrow 2 \mathrm{ae}=2 \sqrt{3} \Rightarrow \mathrm{ae}=\sqrt{3} \\
& \text { and } 2 \mathrm{Ae}^{\prime}=2 \sqrt{3} \Rightarrow \mathrm{Ae}^{\prime}=\sqrt{3} \\
& \Rightarrow \mathrm{ae}=\mathrm{Ae}^{\prime} \Rightarrow \frac{\mathrm{e}}{\mathrm{e}^{\prime}}=\frac{\mathrm{A}}{\mathrm{a}} \\
& \Rightarrow \frac{1}{3}=\frac{\mathrm{A}}{\mathrm{a}} \Rightarrow \mathrm{a}=3 \mathrm{~A}
\end{aligned}
$$
Now $\mathrm{a}-\mathrm{A}=2 \Rightarrow \mathrm{a}-\frac{\mathrm{a}}{3}-2 \Rightarrow \mathrm{a}=3$ and $\mathrm{A}=1$
$$
\begin{aligned}
& \mathrm{Ae}=\sqrt{3} \Rightarrow \mathrm{e}=\frac{1}{\sqrt{3}} \text { and } \mathrm{e}^{\prime}=\sqrt{3} \\
& \mathrm{~b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right) \\
& \mathrm{b}^2=6
\end{aligned}
$$
and $B^2=A^2\left(\left(e^{\prime}\right)^2-1\right)=(2) \Rightarrow B^2=2$
$$
\text { sum of } L R=\frac{2 b^2}{a}+\frac{2 \mathrm{~B}^2}{\mathrm{~A}}=8
$$
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