An ellipse intersects the hyperbola `2x^2-2y^2 =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 (a) the foci of ellipse are `(+-1, 0)` (b) equation of ellipse is `x^2+ 2y^2 =2` (c) the foci of ellipse are `(pmsqrt2,0)`(d) equation of ellipse is `x^2+ 2y^2 =1`

Options

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)`

Answer

A, B

Explanation

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`