Q.

If a>b and e is the eccentricity of the ellipse x2a2+y2b2=1 then the equation of the normal
at the end of the latusrectum in the first quadrant is

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a

x-ey+ae3=0

b

x+ey-ae3=0

c

x+ey+ae3=0

d

x-ey-ae3=0

answer is D.

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Detailed Solution

Given ellipse x2a2+y2b2=1 (a>b)

one end of latusrectum (L)=ae, b2a in Q1

Equation of normal at 'L' is 

a2xae-b2yb2a=a2-b2  axe-ay=a2-a2(1-e2)  axe-y=a21-1+e2  x-eye=a(e)2  x-ey=ae3  x-ey-ae3=0

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