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Question

If e is eccentricity of ellipse x2a2+y2b2=1(a>b) and e is eccentricity of x2a2+y2b2=1(a<b) then

A
e=e
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B
ee=1
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C
1e2+1(e)2=1
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D
None of these
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Solution

The correct option is C 1e2+1(e)2=1
We have,
e=1b2a2(b<a)e=1a2b2(a>b)
Putting the value of e and e in equation.
1e2+1(e)2=111b2a2+11a2b2=1a2a2b2b2a2b2=1
L.H.S=R.H.S
Hence, proved
Option C is correct answer.

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