Let Δ1 be the area of a triangle PQR inscribed in an ellipse and Δ2 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 437 then the ratio 7Δ22Δ1=…..

# 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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### Solution:

${\mathrm{P}}^{\mathrm{\prime }}\left(\mathrm{acos}{\mathrm{\theta }}_{1},\mathrm{asin}{\mathrm{\theta }}_{1}\right){\mathrm{Q}}^{\mathrm{\prime }}\left(\mathrm{acos}{\mathrm{\theta }}_{2},\mathrm{asin}{\mathrm{\theta }}_{2}\right)$ 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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