Normal Latus Rectum point,end of minor axis

Question

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.

Point of lactus rectum are correct, put it in the ellipse equation and equation of eccentricity is derieved — Samar Imam Zaidi, Nov 12 '17 at 09:17

score 1 · Accepted Answer · answered Nov 12 '17 at 09:25

1

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 $(\pm b,0)$

answered Nov 12 '17 at 09:25

lab bhattacharjee

274,582

Normal Latus Rectum point,end of minor axis

1 Answers1