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An ellipse intersects the hyperbola 2x^2...

An ellipse intersects the hyperbola `2x^2-2y =1` orthogonally. The eccentricity of the ellipse is reciprocal to that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then (b) the foci of ellipse are `(+-1, 0)` (a) equation of ellipse is `x^2+ 2y^2 =2` (d) the foci of ellipse are `(t 2, 0)` (c) equation of ellipse is `(x^2 2y)`

A

the equation of the ellipse is `x^(2)+2y^(2)=1`

B

the foci of the ellipse are `(pm1,0)`

C

the equation of the ellipse is `x^(2)+2y^(2)=4`

D

the foci of the ellipse are `(pmsqrt2,0)`

Text Solution

Verified by Experts

The correct Answer is:
A, B
Let the ellipse be
`(x^(2))/(a^(2))+(y^(2))/(b^(2))=1`.
Since ellipse intersects hyperbola orthogonally. The ellipse and hyperbola will be confocal. So, comparing coordinates of foci
`(pma+(1)/(sqrt2),0)=(pm1,0)`
`"or "a=sqrt2 and e=(1)/(sqrt2)`
Now, `b^(2)=a^(2)(1-e^(2))`
`"or "b^(2)=1`
Therefore, the equation of the ellipse is
`(x^(2))/(2)+(y^(2))/(1)=1`
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