The equation of the ellipse , with axes parallel to the coordinates axes , whose eccentricity is `(1)/(3)` and foci at (2,-2) and (2,4) is

Options

A. `((x-1)^(2))/(8)+((Y-2)^(2))/(9)=9`
B. `((x-2)^(2))/(8)+((Y-1)^(2))/(9)=9`
C. `((x-1)^(2))/(9)+((Y-2)^(2))/(8)=9`
D. `((x-2)^(2))/(9)+((Y-2)^(2))/(8)=9`

Answer

B

Explanation

the centre of the ellipse is at the mid point of its foci . So coordinates of centre are (2,1)
clearly , two foci lie on the line x=2 So, major axis is parallel to y-axis .
the distance between two foci is 6.
`therefore 2be =6implies (2b)/(3)=6implies b=9[:' e=(1)/(2)(" given")]`
Now,` a^(2)=b^(2)(1-e^(2))implies a^(2)=81(1-(1)/(9) )=72`
hence , the equation of the ellipse is `((x-2)^(2))/(72)+((y-1)^(2))/(81)=1`.