Question

A parabola is drawn whose focus is one of the foci of the ellipse `x^2/a^2 + y^2/b^2 = 1` (where a>b) and whose directrix passes through the other focus and perpendicular to the major axes of the ellipse. Then the eccentricity of the ellipse for which the length of latus-rectum of the ellipse and the parabola are same is

Answer 1

`x^2/a^2+y^2/b^2=1`
focus of parabola=(ae,0)
Directrix=X-ae
X=-A
`y^2=4AX`
A=ae
Lactus rectum of parabola=Lactus rectum of ellipse
`4A=(2b)^2/a`
`4ae=(2b)^2/a`
`e=b^2/(2a^2)`
We have,
`e^2=(1-b^2)/a^2=1-2e`
`e^2+2e-1=0`
`e=(-2pmsqrt(4+2))/2`
`e=-1pmsqrt2`
`e=-1+sqrt2`