Consider the ellipse `E_1, x^2/a^2+y^2/b^2=1,(a>b)`. An ellipse `E_2` passes through the extremities of the major axis of `E_1` and has its foci at the ends of its minor axis.Consider the following property:Sum of focal distances of any point on an ellipse is equal to its major axis. Equation of `E_2` is

Options

A. `2:1`
B. `3:2`
C. `sqrt2:1`
D. `5:2`

Answer

D

Explanation

We have
`b^(2)=a^(2)(1-e_(1)^(2))`
`"and "2be^(2)=2arArre_(2)=(a)/(b)`
`"So, "(1)/(e_(2)^(2))=1-e_(1)^(2)`
`rArr" "e_(2)^(2)(1-e_(1)^(2))=1`
Tangent at point P `(a cos theta, b sin theta)` on the ellipse is
`(x)/(a) cos theta+(y)/(b) sin theta=1`
It passes through `(0, be_(2))`.
`"So, "e_(2) sin theta=1`
`rArr" "sin theta=(1)/(e_(2))`
`therefore" "theta=tan^(2)((1)/(sqrt(e_(2)^(2)-1)))`
`A_(1)=4xx(1)/(2)xxae_(1)xxbe_(2)=2abe_(1)e_(2)`
`A_(2)=((2a)/(e_(1)))((2b)/(e_(2)))=(4ab)/(e_(1)e_(2))`
`(A_(1))/(A_(2))=(e_(1)^(2)e_(2)^(2))/(2)=2`
`rArr" "e_(1)e_(2)=2`
`"But "e_(2)^(2)(1-e_(1)^(2))=1`
`"So, "e_(2)^(2)-4=1`
`therefore" "e_(2)=sqrt5`
`"and "e_(1)=(2)/(sqrt5)`
`therefore" "e_(2):e_(1)=5:2`