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

Options

A. equation of ellipse is `x^(2)=2y^(2)=2`
B. the foci of ellipse are `(pm 1, 0)`
C. equation of ellipase is `x^(2)+2y^(2)=4`
D. the foci of ellipse are `(pm sqrt(2), 0)`

Explanation

Given `2x^(2)-2y^(2)=1`
`rArr (x^(2))/(((1)/(2)))-(y^(2))/(((1)/(2)))=1 " ...(i)" `
Eccentricity of hyperbola `=sqrt(2)`
So, eccentricity of ellipse `=1//sqrt(2)`
Let equation of ellipse be
`(x^(2))/(a^(2))+(y^(2))/(b^(2))=1 " " ` [where a gt b]
`therefore (1)/(sqrt(2))=sqrt(1-(b^(2))/(a^(2)))`
`rArr(b^(2))/(a^(2))=(1)/(2)rArr a^(2)=2b^(2)`
`rArr x^(2)+2y^(2)+2b^(2) " ...(ii)" `
Let ellipse and hyperbola inersect intersect at
`A((1)/(sqrt(2))sec theta, (1)/(sqrt(2))tan theta)`
On differentiating Eq. (i), we get
`4x-4y(dy)/(dx)=0 rArr (dy)/(dx)=(x)/(y)`
`(dy)/(dx)"|"_(at A) =(sec theta )/(tan theta)="cosec"theta`
Since, ellipse and hyperbola are orthogonal.
` therefore -(1)/(2) "cosec"^(2)theta= -1 rArr "cosec"^(2)theta=2 rArr theta = pm (pi)/(4)`
`therefore A(1,(1)/(sqrt(2))) or (1,-(1)/(sqrt(2)))`
From Eq. (ii), `1+2((1)/(sqrt(2)))^(2)=2b^(2)`
`rArr b^(2) =1`
Equation of ellips is `x^(2)+2y^(2)=2.`
Coordinates of foci `(pm ae, 0)=(pm sqrt(2) * (1)/(sqrt(2)),0)=(pm 1, 0)`
If major axis is along Y-axis, then
`(1)/(sqrt(2))=sqrt(1-(a^(2))/(b^(2))) rArr b^(2)=2a^(2)`
` therefore 2x^(2)+y^(2)=2a^(2) rArr Y'= -(2x)/(y)`
`rArr y'_(((1)/(sqrt(2))sec theta","(1)/(sqrt(2)) tan theta))=(-2)/(sin theta)`
As ellipse and hyperbola are orthogonal
` therefore -(2)/(sin theta)*"cosec"theta= -1`
`rArr "cosec"^(2)theta=(1)/(2) rArr theta = pm (pi)/(4)`
` therefore 2x^(2)+y^(2)=2a^(2)`
`rArr 2+(1)/(2)=2a^(2) rArr a^(2)=(5)/(4)`
`therefore 2x^(2)+y^(2)=(5)/(2),` corresponding foci are `(0, pm1)`.
Hence, option (a) and (b) are correct.