The foci of the ellipse `((x-3)^(2))/(36)+((y+2)^(2))/(16)=1,` are

Options

A. `(3+-2sqrt(5),2)`
B. `(3+-2sqrt(5),-2)`
C. `(3-2)`
D. none of these

Answer

B

Explanation

shifiting the orgin at `(3,-2)` we have
`x=X+3 and y=Y-2`
the equation of the ellipse with reference to new axes is
`(X^(2))/(36)+(Y^(2))/(16)=1`
comparing it with `(X^(2))/(a^(2))+(Y^(2))/(b^(2))=1` we get `a^(2)=36 and b^(2)=16`
Let e be the eccentricity of this ellipse , then ,
`e=sqrt(1-(b^(2))/(a^(2)))impliese=sqrt(1-(16)/(36))=(sqrt(5))/(3)`
the coordinates of the foci with reference to new axes are
`(X=+- ae,Y=0) or (X=+- 2sqrt(5),Y=0).`
subsituting these in (i) , we obtain the coordinates of foci as
`(3+- 2sqrt(5),-2)`