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Question

Consider an ellipse whose focus is at (ae,0) and directrix along (x=ae).
Find the equation of the ellipse if

b2=a2(1e2)


A

x2a2+y2b2=1

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B

x2a2y2b2=1

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C

x2a2+y2b2=0

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D

x2a2+y2b2=1

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Solution

The correct option is A

x2a2+y2b2=1


As per definition of ellipse, Let the point be S(h, k), focus be (p,q) and directrix be lm+my+n=0
Distance between s &(p,q)Distance from lx+my+n=0=e(hp)2+(kq)2=e2(hl+mk+nll2+m2)2(l2+m2)[(hp)2+(kq)2]=e2(lh+mk+n)2Here l=1, m=0, n=ae, p=ae, q=0[(hae)2+(k0)2]=e2(lh+(0)k+(ae))2h2+k2=(eh)2+a2(1e2)h2(a2)+k2(a2)(1e2)=1will set b2=a2(1e2) x2a2+y2b2=1


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