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If s, s' are the length of the perpendicular on a tangent from  the foci, a,a' are those from the vertices, c is that from the  centre and e is the eceentricity of the ellipse, x2a2+y2b2=1 ,  then ss'-c2aa'-c2=

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a

e

b

1e

c

1e2

d

e2

answer is D.

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Detailed Solution

Let the equation of tangent y=mx+a2 m2+b2 Foci ≡(± ae ,0), vertices ≡(±a,0),C≡(0,0)∴s=mae+a2 m2+b21+m2,s'=-mae+a2 m2+b21+m2a=ma+a2m2+b21+m2,a'=-ma+a2m2+b21+m2, c=a2m2+b21+m2∴ss'-c2aa'-c2=-m2a2e21+m2-m2a21+m2=e2
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If s, s' are the length of the perpendicular on a tangent from  the foci, a,a' are those from the vertices, c is that from the  centre and e is the eceentricity of the ellipse, x2a2+y2b2=1 ,  then ss'-c2aa'-c2=