Question

If the origin is shifted to the point $(\frac{ab}{a-b},0)$ without rotation, then the equation $(a-b)(x^2+y^2)-2abx=0$ becomes

• 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. $(a-b{)}^2(X^2+Y^2)=a^2{b}^2$

Answer 1

The correct option is D $(a-b{)}^2(X^2+Y^2)=a^2{b}^2$

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 a new point (X, Y) with reference to new axes, such that
$x=X+\frac{ab}{a-b},y=Y+0$
substituting these in (i), we get
$(a-b)[{(X+\frac{ab}{a-b})}^2+y^2]-2ab[X+\frac{ab}{a-b}]=0$
$\Rightarrow (a-b)[X^2+\frac{a^2{b}^2}{(a-b{)}^2}+Y^2+\frac{2abX}{a-b}]-2abX-\frac{2a^2{b}^2}{a-b}=0$
$\Rightarrow (a-b)(X^2+Y^2)=\frac{a^2{b}^2}{a-b}$
$\Rightarrow (a-b{)}^2(X^2+Y^2)=a^2{b}^2$