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  • Let S and  be the foci of an ellipse and B be any one of the

Let S and \mathrm{S^{\prime}} be the foci of an ellipse and B be any one of the extremities of its minor axis. If \mathrm{\triangle S^{\prime} B S} is a right angled triangle with right angle at B and area \mathrm{\left(\triangle S^{\prime} B S\right)=8} sq. units, then the length of a latus rectum of the ellipse is

Option: 1

1


Option: 2

2


Option: 3

3


Option: 4

4


Answers (1)

best_answer

Since \mathrm{\angle S^{\prime} B S=90^{\circ}}

\therefore  Slope of \mathrm{S^{\prime} B \times \, \, Slope \, \, of \, \, S B=-1}

\mathrm{ \Rightarrow \quad \frac{-b}{-a e} \times \frac{-b}{a e}=-1 \Rightarrow b^2=a^2 e^2 }

Now, area of \mathrm{\triangle S^{\prime} B S=8}
\mathrm{ \begin{aligned} & \Rightarrow \quad \frac{1}{2} \times 2 a e \times b=8 \\\\ & \Rightarrow \quad a e \times b=8 \\\\ & \Rightarrow \quad a^2 e^2 \times b^2=64 \\\\ & \Rightarrow \quad\left(a^2 e^2\right)^2=64 \\\\ & \Rightarrow \quad a^2 e^2=8 \Rightarrow a e=2 \sqrt{2} \\\\ & \therefore \quad b=2 \sqrt{2} \\\\ & \text { Again, } a^2=a^2 e^2+b^2 \\\\ & \Rightarrow \quad a^2=8+8=16\, \, \, \, \, \, \, \, \, \left[\because e=\sqrt{1-\frac{b^2}{a^2}}\right] \\\\ & \Rightarrow \quad a=4 \end{aligned} }

\mathrm{\therefore \text { Length of latus rectum }=\frac{2 b^2}{a}=\frac{2 \times 8}{4}=4}

 

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jitender.kumar

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