prove that the equation of eccentricity of an ellipse can be obtained by equation e^4+e^2-1=0 if the normal drawn from one end of latus rectum of an ellipse passes through other end of the minor axis

Let the equation of ellipse be x^2/a^2 + y^2/b^2 =1Slope of normal to ellipse at (x1,y1)= (a^2*y1)/(b^2*x1)Slope of normal at L(ae,b^2/a)=1/e (using above equation)equation of normal to ellipse at L is (ax/e) -ay=a^2 - b^2 (using y-y1=m(x-x1))Since it passes through B(0,-b), substituting in above equation of normal, we getab =a^2 - b^2And b= a*sqrt(1-e^2)finally we get sqrt(1-e^2) = e^2Squaring both sides, we get e^4 + e^2 -1= 0