Question

# If the normal at the end of latus rectum of an ellipse x2a2+y2b2=1 of eccentricity e passes through one end of the minor axis then e4+e2=

Solution

## The correct option is B  Latus rectum =(ae,b2a) Equation of normal is a2xae−b2yb2a=a2−b2⇒axe−ay=a2−b2 Since it passes through (0, b) then −ab=a2−b2⇒√1−e2=e2⇒1−e2=e4

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