If the origin is shifted to the point `((a b)/(a-b),0)` without rotation, then the equation `(a-b)(x^2+y^2)-2a b x=0` becomes

Options

A. A. `(a-b)(x^2+y^2)-(a+b)xy+abx=a^2`
B. B. `(a+b)(x^2+y^2)=2ab`
C. C. `(x^2+y^2)=(a^2+b^2)`
D. D. `(a-b)^2(x^2+y^2)=a^2b^2`

Answer

D

Explanation

The given equation is
`(a-b)(x^2+y)-2abx=0`
The origin is shifted to `(ab//a-b),0)` Any point `(x,y)` on the curve (1) must be replaced with a new point (X,Y) with reference to the new axes, such that
`x=X+(ab)/(a-b)and y=Y+0`
Substituting these in (i), we get
`(a-b)[(X+(ab)/(a-b))^(2) +Y^2]-2ab[X+(ab)/(a-b)]=0`
or `(a-b)[X^(2)+(a^(2)b^(2))/(a-b)^(2)+Y^2+(2abX)/(a-b)]-2abX-(2a^2b^2)/(a-b)=0`
or `(a-b)(X^2+Y^2)=(a^2b^2)/(a-b)`
or `(a-b)^(2)(X^2+Y^2)=a^2b^2`