Find the equation of the ellipse whose axes are of length 6 and `2sqrt(6)` and their equations are `x-3y+3=0` and `3x+y-1=0` , respectively.

Options

A. `2(x-3y+3)^(2)+3(3x+y-1)^(2)= 180`
B. `3(x-3y+3)^(2)+2(3x+y-1)^(2)= 180`
C. `2(x-3y+3)^(2)+3(3x+y-1)^(2)= 18`
D. `3(x-3y+3)^(2)+2(3x+y-1)^(2)= 18`

Answer

B

Explanation

Let P(x,y) be any point on the ellipse and let `P_(1) and P_(2)` be the length of perpendiculars drawn from P on the major and minor of the ellipse . Then ,
`P_(1)=(x-3y+3)/(sqrt(1+9)) and P_(2) =(3x+y-1)/(sqrt(9+1))`
let 2a be the length of major and minor axes of the ellipse.
we have ,
`2a=6 and 2b=2sqrt(6) implies a=3 and b=sqrt(6)`
the equation of the ellipse is
`(P_(1)^(2))/(b^(2))+(P_(2)^(2))/(a^(2))=1`
`implies {((w-3y+3)/(sqrt(1+9)))/(sqrt(6))}^(2)+{((3x+y-1)/(sqrt(9+1)))/(3)}^(2)=1`
`implies ((x-3y+3)^(2))/(60)+((3x+y-1)^(2))/(90)=1`
`implies3(x-3y+3)^(2)+2(3x+y-1)^(2)=180`