Question:

For the ellipse \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \] take a point \[ (a\cos\theta,b\sin\theta). \] The normal at this point is \[ a x\sec\theta-b y\cosec\theta=a^2-b^2. \] At an end of the latus rectum, \[ x=ae. \] Thus \[ a\cos\theta=ae \quad\Longrightarrow\quad \cos\theta=e. \] Also, \[ \sin\theta=\sqrt{1-e^2}=\frac{b}{a}. \] So the normal becomes \[ \frac{a}{e}x-a y=a^2-b^2. \] Since \[ a^2-b^2=a^2e^2, \] we get \[ \frac{a}{e}x-a y=a^2e^2. \] This normal passes through one end of the minor axis. Taking \((0,-b)\), \[ 0+a b=a^2e^2. \] Hence \[ \frac{b}{a}=e^2. \] But \[ \frac{b}{a}=\sqrt{1-e^2}. \] Therefore \[ \sqrt{1-e^2}=e^2. \] Squaring, \[ 1-e^2=e^4. \] Thus \[ e^4+e^2-1=0. \] \[ \boxed{e^4+e^2-1=0} \]