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# Let ${\mathrm{\Delta }}_{1}$ be the area of a triangle PQR inscribed in an ellipse and ${\mathrm{\Delta }}_{2}$ be the area of the triangle P'Q'R' whose vertices are the points lying on the auxiliary circle corresponding to the points P, Q, R respectively. If the eccentricity of the ellipse is $\frac{4\sqrt{3}}{7}$ then the ratio $\frac{7{\mathrm{\Delta }}_{2}}{2{\mathrm{\Delta }}_{1}}=$…..

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detailed solution

3.5

${\mathrm{P}}^{\mathrm{\prime }}\left(\mathrm{acos}{\mathrm{\theta }}_{1},\mathrm{asin}{\mathrm{\theta }}_{1}\right){\mathrm{Q}}^{\mathrm{\prime }}\left(\mathrm{acos}{\mathrm{\theta }}_{2},\mathrm{asin}{\mathrm{\theta }}_{2}\right)$ and ${\mathrm{R}}^{\mathrm{\prime }}\left(\mathrm{acos}{\mathrm{\theta }}_{3},\mathrm{asin}{\mathrm{\theta }}_{3}\right)$
Clearly $\frac{{\mathrm{\Delta }}_{1}}{{\mathrm{\Delta }}_{2}}=\frac{\mathrm{b}}{\mathrm{a}}=\sqrt{1-{\mathrm{e}}^{2}}=\frac{1}{7}$
$\therefore \frac{7{\mathrm{\Delta }}_{2}}{2{\mathrm{\Delta }}_{1}}=3.5$

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