Q.29. Ellipse

Question 29.
If the normal at the end of  latus rectum of the ellipse x^2/a^2 + y^2/b^2 = 1 passes through (0,-b), then
e^4 + e^2  equals         
a) 1          b) √2          c) (√(5)-1)/2          d) (√(5)+1)/2

Answer 29.
x^2/a^2 + y^2/b^2 = 1
=> 2x/a^2 + 2y/b^2 * dy/dx = 0
=> dy/dx = - b^2*x/a^2*y
=> slope of normal
= - dx/dy = a^2*y/b^2*x

 End of the latus rectum of ellipse is (ae, b^2/a)
=> slope of normal at the endpoint of the latus rectum
= a^2 * (b^2/a) / b^2 * ae
= 1/e
=> eqn. of normal at the endpoint of latus rectum is
y - b^2/a = (1/e)(x - ae)

 If it passes gthrough (0, - b),
- b - b^2/a = (1/e) (0 - ae)
=> b + b^2/a = a
=> ab + b^2 = a^2
=> ab = a^2 - b^2
=> b/a = a^2 - b^2/a^2
=> √(1 - e^2) = e^2
=> 1 - e^2 = e^4
=> e^4 + e^2 = 1
=> Answer ia a) 1.

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