Find the equation of normals at the end of latus rectum,and prove that each passes through each passes through an end of the minor axis if $e^4+e^2=1$.
My approach , as the word minor axis is given by default it is ellipse. Now equation of ellipse is $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ The point at the end of latus rectum are $(ae,\frac{b^2}{a})$&$(ae,-\frac{b^2}{a})$ & points at the end of minor axis are (0,b)&(0,-b) . I tried to equate that slope are equal but not able to do so.
Hint:
$b^2=a^2(1-e^2),b^2/a=?$
Use http://www.askiitians.com/iit-jee-coordinate-geometry/tangent-and-normal.aspx to find the equation of normal at $(ae,b^2/a),$
Now this has to pass through $(\pm b,0)$