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