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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.

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  • $\begingroup$ Haha rectum point. $\endgroup$ – Oria Gruber Nov 12 '17 at 9:09
  • $\begingroup$ Point of lactus rectum are correct, put it in the ellipse equation and equation of eccentricity is derieved $\endgroup$ – Samar Imam Zaidi Nov 12 '17 at 9:17
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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 $(\pm b,0)$

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