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A parabola is drawn with focus at one of...

A parabola is drawn with focus at one of the foci of the ellipse `(x^2)/(a^2)+(y^2)/(b^2)=1` . If the latus rectum of the ellipse and that of the parabola are same, then the eccentricity of the ellipse is (a) `1-1/(sqrt(2))` (b) `2sqrt(2)-2` (c) `sqrt(2)-1` (d) none of these

A

`1-(1)/(sqrt(2))`

B

`2sqrt(2)-2`

C

`sqrt(2)-1`

D

none of these

Text Solution

Verified by Experts

(3) The equation of the ellipse is `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1`
The equation of the parabola with S)ae,0) and directrix x+ae=o is `y^(2)=4aex`.
Now, the length of latus rectum of the ellipse is `2b^(2)//a` and that of the parabola is 4ae.

For the latus recta tot be equal , we get
`(2b^(2))/(a=4ae`
or `(2a^(2)(1-e^(2)))/(a)=4ae`
`or 1-e^(2)=2e`
or `e^(2)2e-1=0`
Therefore, `e=(-2+-sqrt(8))/(2)=+-sqrt(2)`
Hence, `e=sqrt(2)-1`
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