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Question

Mathematics Question on Ellipse

Let SS and SS' be the foci of an ellipse and B be one end of its minor axis. If SBSSBS' is an isosceles right angled triangle then the eccentricity of the ellipse is

A

12\frac{1}{\sqrt{2}}

B

12\frac{1}{2}

C

32\frac{\sqrt{3}}{2}

D

13\frac{1}{3}

Answer

12\frac{1}{\sqrt{2}}

Explanation

Solution

We have,
SBSS B S' is an isosceles right angle triangle.

SS2=SB2+SB2\therefore S S^{2}=S B^{2}+S^{\prime} B^{2}
(2ae)2=b2+(ae2+b2+(ae)2\Rightarrow (2 \,a e)^{2}=b^{2}+\left(a e^{2}+b^{2}+(a e)^{2}\right.
4(ae)2=2(b2+(ae)2)\Rightarrow 4(a e)^{2}=2\left(b^{2}+(a e)^{2}\right)
(ae)2=b2\Rightarrow (a e)^{2}=b^{2}
e2=b2a2\Rightarrow e^{2}=\frac{b^{2}}{a^{2}}
1e2=1b2a2\Rightarrow 1-e^{2}=1-\frac{b^{2}}{a^{2}}
1e2=e2[e=1b2a2]\Rightarrow 1-e^{2}=e^{2} \left[\because e=\sqrt{1-\frac{b^{2}}{a^{2}}}\right]
2e2=1\Rightarrow 2e^{2}=1
e=12\Rightarrow e = \frac{1}{\sqrt{2}}