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-b)(X^2+Y^2)-(a+b)XY+abX=a^2`
B. `(a+b)(X^2+Y^2)=2ab`
C. `(X^2+Y^2)=(a^2+b^2)`
D. `(a-b)^2(X^2+Y^2)=a^2b^2`

Answer

D

Explanation

The given equation is `(a-b)(x^2+y^2)=-2abx=0` ....(i)
The origin is shifted to `(ab//(a-b),0)`. Any point `(x,y)` on the curve (i) must be replaced with new point `(X,Y)` with reference to new axes , such that `x=X+(ab)/(a-b)and y=Y+0`
On substituting in Eq.(i), we get `(a-b)[(X+(ab)/(a-b))^2+Y^2]-2ab[X+(ab)/(a-b)]=0`
`rArr (a-b)[X^2+(a^2b^2)/((a-b))^2+Y^2+(2abX)/(a-b)]-2abX-(2a^2b^2)/(a-b)=0`
`rArr ((bg-fh)/(h^2-ab),(af-gh)/(h^2-ab))`
`rArr (a-b)(X^2+Y^2)=(a^2b^2)/(a-b)`
`therefore (a-b)^2(X^2+Y^2)=a^2b^2` .