If two concentric ellipses of eccentricities e,e1 are such that the foci of one of ellipse lies on other , and their length of major axes are equal and S1: If the angle between their major axes of ellipse is θ then sinθ=1−1e21−1e12S2: The quadrilateral formed by joining the foci of two ellipse is a parallelogram, then

Diagonals of quadrilateral H S H1 S1 bisect each other, so it is parallelogram Let OH= ae1⇒H=ae1cos⁡θ,ae1sin⁡θH lies on x2a2+y2a21−e2=1 sin2⁡θ=1−1e21−1e12

If two concentric ellipses of eccentricities e,e1 are such that the foci of one of ellipse lies on other , and their length of major axes are equal and S1: If the angle between their major axes of ellipse is θ then sinθ=1−1e21−1e12S2: The quadrilateral formed by joining the foci of two ellipse is a parallelogram, then

Diagonals of quadrilateral H S H1 S1 bisect each other, so it is parallelogram Let OH= ae1⇒H=ae1cos⁡θ,ae1sin⁡θH lies on x2a2+y2a21−e2=1 sin2⁡θ=1−1e21−1e12

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If two concentric ellipses of eccentricities e,e¹ are such that the foci of one of ellipse lies on other , and their length of major axes are equal and
S₁: If the angle between their major axes of ellipse is $θ$ then $sinθ = \sqrt{(1 - \frac{1}{e^{2}}) (1 - \frac{1}{e^{1^{2}}})}$
S₂: The quadrilateral formed by joining the foci of two ellipse is a parallelogram, then

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S₁ and S₂are true

S₁ is true, S₂ is false

S₁ is false, and S₂is true

S₁ and S₂ are false

answer is A.

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Detailed Solution

Diagonals of quadrilateral H S H¹ S¹bisect each other, so it is parallelogram Let OH=
${ae}^{1} \Rightarrow H = ({ae}^{1} \cos θ, {ae}^{1} \sin θ)$
$H lies on \frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2} (1 - e^{2})} = 1 \sin^{2} θ = (1 - \frac{1}{e^{2}}) (1 - \frac{1}{e^{1^{2}}})$
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