The foci of the conic `25x^(2) +16y^(2)-150 x=175` are :

Options

A. `(0,+-3)`
B. `(0,+- 2)`
C. `(3,+- 3)`
D. `(0,+-1)`

Answer

C

Explanation

the given conic can be re - written as
`25(x^(2)-6x+9)+16y^(2)=400`
`implies ((x-3)^(2))/(4^(2))+((y-0)^(2))/(5^(2))=1`
shifing the origin at(3,0) without rotating the coordinate axes , we have
`x=X+3, y=Y+0`
the equation of the ellipse with reference to new axes is
`(X^(2))/(4^(2))+(y^(2))/(5^(2))=1`
comparing this equation with the standard equation
`(X^(2))/(a^(2))+(Y^(2))/(b^(2))=1`, we get `a=4,b=5,`
Let e be eccentricity of the given conic , then ,
`e=sqrt(1-(a^(2))/(b^(2)))=sqrt(1)-(16)/(25)=(3)/(5)`
the coordinates of foci with respect to new origin are
`(X=0, Y=+-be)=(X=0,Y=+-3)`
Substituting these in (i) we obtain `(3,+- 3)` as the coordinates of foci.