Find the equation of an ellipse hose axes lie along the coordinate axes, which passes through the point (-3,1) and has eccentricity equal to `sqrt(2//5)dot`

Options

A. `3x^(2)+5y^(2)-32=0`
B. `5x^(2)+3y^(2)-48-0`
C. `3x^(2)+5y^(2)-15=0`
D. `5x^(2)+3y^(2)-32=0`

Answer

D

Explanation

Let the equation of the ellipse be `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1 and ` let e be its eccentricity .
it passes through `(-3,1)` and has eccentricity `e=sqrt(2//5)`
`therefore (9)/(a^(2))+(1)/(b^(2))=1 and e^(2)=(2)/(5)`
`implies (9)/(a^(2))+(1)/((1-e^(2)))=1 and e^(2)=(2)/(5)`
` implies (9)/(a^(2))+(5)/(3a^(2))=1 implies b^(2)=(32)/(3)(1-(2)/(5))=(32)/(5)`
hence . the ellipse` 5x^(2)+3y^(2)=32`