Let $Δ_{1}$ be the area of a triangle PQR inscribed in an ellipse and $Δ_{2}$ 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 $\frac{7 Δ_{2}}{2 Δ_{1}} =$ …..

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answer is 3.5.

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Detailed Solution

$P^{'} (acos θ_{1}, asin θ_{1}) Q^{'} (acos θ_{2}, asin θ_{2})$ and $R^{'} (acos θ_{3}, asin θ_{3})$
Clearly $\frac{Δ_{1}}{Δ_{2}} = \frac{b}{a} = \sqrt{1 - e^{2}} = \frac{1}{7}$
$∴ \frac{7 Δ_{2}}{2 Δ_{1}} = 3 .5$

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