Let `x^(2)=4ky,k gt 0` be a parabola with vertex A. Let BC be its latus rectum. An ellipse with center on BC touches the parabola at A, and cuts BC at point D and E such that BD=DE=EC (B,D,E,C in that order). The eccentricity of the ellipse is

Options

A. `(1)/(sqrt(2))`
B. `(1)/(sqrt(3))`
C. `(sqrt(5))/(3)`
D. `(sqrt(3))/(2)`

Answer

C

Explanation

n For the parabola,
verter A(0,0)
Four F:(0,k)
end point of latus rectum:
B(-2k,k): c(2k,k)
Length of BC = 4k,
BD=DE=EC
And BD+DE+EC=4k
1`impliesBD=DE=EC=(4k)/(3)`…….(i)

So major Axis of ellipse =2AF= `2xx` minor Axis of Elllipse =`DE=(4k)/(3)`
Eccentricity =`C=sqrt(1-(b^(2))/(a_(2))=sqrt((1-((2k)/(3))^(2))/(k^(2)))" " 2=(sqrt(5))/(3)`