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If two concentric ellipse be such that the foci of one be on the other and if 3 / 5 and 4 / 5  be their eccentricities. If \theta  be the angle between their axes, then the value of \mathrm{1+\sin \theta+\sin ^4 \theta} must be

Option: 1

1


Option: 2

2


Option: 3

3


Option: 4

4


Answers (1)

Let S and \mathrm{S^{\prime}} be the foci of one ellipse and H and \mathrm{H^{\prime}} be the other, C being their common centre. Then, \mathrm{S H S^{\prime} H^{\prime}} is a parallelogram and since
Since the sum of the focal distance of any point on an ellipse is equal to its major axis which is 2a.

Then, \mathrm{C S=a e, C H=a e^{\prime}}

Let \theta be the angle between their axes.

Then, \mathrm{S H^2=a^2 e^2+a^2 e^2-2 a^2 e e^{\prime} \cos \theta}

\mathrm{ H S^{\prime 2}=a^2 e^2+a^2 e^{\prime 2}+2 a^2 e e^{\prime} \cos \theta }

Now, \mathrm{2 a=S H+S^{\prime} H}

Squaring both sides, then

\mathrm{ \begin{aligned} & 4 a^2=(S H)^2+\left(S^{\prime} H\right)^2+2(S H) \cdot\left(S^{\prime} H\right) \\\\ & \Rightarrow 4 a^2=2 a^2\left(e^2+e^{\prime 2}\right)+2 \sqrt{\left(a^2 e^2+a^2 e^{\prime 2}\right)^2-4 a^4 e^2 e^{\prime 2} \cos ^2 \theta} \\\\ & \Rightarrow \quad\left(2-e^2-e^{\prime 2}\right)^2=\left(e^2+e^{\prime 2}\right)^2-4 e^2 e^{\prime 2} \cos ^2 \theta \\\\ & \Rightarrow \quad 4+\left(e^2+e^{\prime 2}\right)^2-4\left(e^2+e^{\prime 2}\right)=\left(e^2-e^{\prime 2}\right)^2-4 e^{2 / 2} \cos ^2 \theta \\\\ & \Rightarrow \quad 1-e^2-e^{\prime}=-e^2 e^{\prime} \cos ^2 \theta \\\\ & \therefore \quad \cos \theta=\frac{\sqrt{e^2+e^{\prime 2}-1}}{e e^{\prime}}=\frac{\sqrt{\left(\frac{9}{25}+\frac{16}{25}-1\right)}}{15 / 25}=0 \\ & \therefore \quad \theta=\pi / 2 \\\\ & \therefore \quad \sin \theta=1 \\\\ & \therefore \quad 1+\sin \theta+\sin ^4 \theta=1+1+1=3 \end{aligned} }

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Ramraj Saini

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