Answer to Question #91741 in Analytic Geometry for Ra

Question

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

Answer 1

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