Let P and Pi1 be the feet of the perpendiculars drawn from the foci S and S1 on a tangent Ti to an ellipse whose length of semi-major axis is 20. If ∑i=110(SPi)(S1Pi1) = 2560, then the value of eccentricity is

∑i=110 SPiS′Pi′=2560 ⇒ 10b2=2560 ⇒ b2=256 ⇒ b=16 ⇒256=4001−e2 1−e2=1625 or e=35

Let P and Pi1 be the feet of the perpendiculars drawn from the foci S and S1 on a tangent Ti to an ellipse whose length of semi-major axis is 20. If ∑i=110(SPi)(S1Pi1) = 2560, then the value of eccentricity is

∑i=110 SPiS′Pi′=2560 ⇒ 10b2=2560 ⇒ b2=256 ⇒ b=16 ⇒256=4001−e2 1−e2=1625 or e=35

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Let $P and P_{i}^{1}$ be the feet of the perpendiculars drawn from the foci S and S¹ on a tangent T_i to an ellipse whose length of semi-major axis is 20. If $\sum_{i = 1}^{10} ({SP}_{i}) (S^{1} P_{i}^{1})$ = 2560, then the value of eccentricity is

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$\sum_{i = 1}^{10} ({SP}_{i}) (S^{'} P_{i}^{'}) = 2560 \Rightarrow 10 b^{2} = 2560 \Rightarrow b^{2} = 256 ⇒ b = 16 ⇒ 256 = 400 (1 - e^{2})$
$1 - e^{2} = \frac{16}{25} or e = \frac{3}{5}$

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Let P and Pi1 be the feet of the perpendiculars drawn from the foci S and S1 on a tangent Ti to an ellipse whose length of semi-major axis is 20. If ∑i=110(SPi)(S1Pi1) = 2560, then the value of eccentricity is

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Let $P and P_{i}^{1}$ be the feet of the perpendiculars drawn from the foci S and S¹ on a tangent T_i to an ellipse whose length of semi-major axis is 20. If $\sum_{i = 1}^{10} ({SP}_{i}) (S^{1} P_{i}^{1})$ = 2560, then the value of eccentricity is