IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

puneet

Heyy Pavneet,

I will answer the first question as the second question is already answered in
http://www.goiit.com/posts/list/1611.htm.

Here goes...

let E(n-1) be x^2/a^2 + y^2/b^2 = 1

and E(n) be x^2/A^2 + y^2/B^2 = 1

Now according to question B = a and b = Ae (since e is constant)

now see that

1 - b^2 / a^2 = e^2 .. (1)

or e^2 = (a^2 - b^2) / a^2 .. (2)

= (A^2 - B^2 ) / A^2

= ( (b/e)^2 - a^2 ) / (b/e)^2 .. (3)

solving (2) and (3) we get

e = (b/a)^2

= 1 - e^2 ( from (1) )

so e^2 + e - 1 = 0

and thus e = (sqrt(5) - 1 )/ 2 .. so answer is (b)

cheers !