Answer to Question #91741 in Analytic Geometry for Ra

Question #91741
If the normal at one extremity of latus rectum of an ellipse passes through one extremity of the minor axis, prove that the eccentricity e is given by e⁴+e²-1=0
1
Expert's answer
2019-07-22T03:48:24-0400

Solution:

General Equation of the Ellipse

x2/a2 + y2/b2 =1.





Let the normal at the extremity P of the latus rectum passes through the extremity D of the minor axis.


The coordinates of P will be (ae,b2 /a ) and the coordinates of D will be (0, −b)


So the equation of the normal at P becomes


( a2x / ae ) - (b2y*a/ b2 ) = a2 - b2

(ax/e ) - ay = a2 - b2  ----------------- (2)


The normal at P passes through D(0, −b) so the equation (2) becomes,


(a*0 /e) - a* (-b) = a2 -b2


0 + ab = a2 - b2


Squaring both the sides


(ab)2 = (a2 - b2)2 ------------------- (3)


We know b2 = a2 (1−e2)


Put the value of b2 in equation (3)


a2 [ a2(1-e2) ] = [a2 - [ a2(1-e2) ] ]2


a4(1-e2) = (a2 - a2 + a2e2)2


a4(1-e2) = a4e4


1-e2 = e4


⟹ 1-e2 - e4 = 0


⟹ e4 + e2 -1 = 0


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