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Find the equation of an ellipse hose axe...

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`

A

`5x^(2)+3y^(2)-32=-0`

B

`3x^(2)+5y^(2)-32=-0`

C

`5x^(2)+3y^(2)-48=-0`

D

`3x^(2)+5y^(2)-15=-0`

Text Solution

Verified by Experts

Le the equation of the ellipse be `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1`
Which will pass through `(-3,1)"if" (9)/(a^(2))+(1)/(b^(2))=1`
And eccentricity `e=sqrt(-(b^(2))/(a^(2)))=sqrt((2)/(3))`
`rArr(2)/(5)=1-(b^(2))/(a^(2))rArr(b^(2))/(a^(2))=(3)/(5)rArrb^(2)=(3)/(5)a^(2)`
Thus `(9)/(a^(2))+(1)/(b^(2))=1 rArr(9)/(a^(2))+(5)/(3a^(2))=1`
`rArra^(2)=(32)/(3),b^(2)=(3)/(5)xx(32)/(3)=(32)/(5)`
Required equation of the ellipse is `3x^(2)+5y^(2)=32`
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