An ellipse intersects the hyperbola `2x^(2)-2y^(2)=1` orthogonally. The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then

Options

A. Equation of the ellipse is `x^(2)+2y^(2)=2` with foci at `(+-1,0)`
B. Equation of the ellipse is `x^(2)+2y^(2)=2` with foci at `(+-sqrt(2),0)`
C. Equation of the ellipse is `x^(2)+2y^(2)=4` with foci at `(+-1,0)`
D. Equation of the ellipse is `x^(2)+2y^(2)=4` with foci at `(+-sqrt(2),0)`

Answer

A

Explanation

Let the equation of the ellipse be `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1`.
We know that the curves `a_(1)x^(2)+b_(1)y^(2)=1` and `a_(2)x^(2)+b_(2)y^(2)=1` intersect orthogonally iff `(1)/(a_(1))-(1)/(b_(1))=(1)/(a_(2))-(1)/(b_(2))`.
Therefore, `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` and `2x^(2)-2y^(2)=1` with intersect orthogonally , if
`a^(2)-b^(2)=(1)/(2)-(-(1)/(2))impliesa^(2)-b^(2)=1`.........`(i)`
Let `e` be the eccentricity of the ellipse. Then,
`e=(1)/(sqrt(2))` [`:'` Eccentricity of the hyperbola `=sqrt(2)`]
`:.b^(2)=a^(2)(1-e^(2))impliesb^(2)=(a^(2))/(2)`.......`(ii)`
From `(i)` and `(ii)`, we get `a^(2)=2`, `b^(2)=1`.
Hence, the equation of the ellipse is
`(x^(2))/(2)+(y^(2))/(1)=1` or , `x^(2)+2y^(2)=2`.
The coordinates of its foci are `(+-1,0)`.