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Question

Let S and S be two foci of the ellipse x2a2+y2b2=1. If a circle described on SS as diameter intersects the ellipse in real and distinct points, then the eccentricity e of the ellipse satisfies

A
e=12
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B
e(12,1)
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C
e(0,12)
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D
None of these
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Solution

The correct option is A e(12,1)

Given equation of ellipse is x2a2+y2b2=1

The foci of the ellipse are S(ae,0) and S(ae,0)

Equation of circle with SS as diameter is

(x+ae)(xae)+(y0)(y0)=0x2a2e2+y2=0x2+y2=a2e2

y2=a2e2x2

Substituting in the equation of ellipse, we have

x2a2+a2e2x2b2=1x2(1a21b2)=1a2e2b2x2(b2a2b2a2)=b2a2e2b2x2(b2a2a2)=b2a2e2x2(b2a21)=a2a2e2a2e2x2(e2)=a2(12e2)x2=a2e2(2e21)x=±ae2e21

Circle intersect the ellipse on distinct points so the values of x will be real and distinct, if

2e21>0e2>12e>±12

But e always lies between (0,1).

e(12,1)


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