Hi sid 178, welcome to PF.

The co-ordinates of the one end of the latus rectum is ( ae, b^2/a).

The slope of the tangent at that point is given by

2x/a^2 + 2yy'/b^2 = 0

y' = [-x1b^2/y1a^2]. Substitute the values of (x1, y1). You get

y' = .......?

Find the slope of the normal. The end point of the minor axis is ( 0, -b).

m = [b^2/a + b]/ae.

Since tangent and normal are perpendicular to each other

m*y' = -1.

Simplify the result by using the relation b^2 = a^2( 1 - e^2).

Answer: e^4 + e^2 - 1 = 0