Question

What does the equation $(a-b)(x^2+y^2)-2abx=0$ become if the origin be moved to the point $(\frac{ab}{a-b},0)$?

Answer 1

$(a-b)(x^2+y^2)-2abx=\mathrm{0..........}\left(1\right)$

Let a point on the curve with respect to origin be $(x,y)$

Let the same point with respect to new origin be $(x^{\prime },y^{\prime })$

The origin is moved from $(0,0)$ to $(\frac{ab}{a-b},0)$

Thus, $x=x^{\prime }+\frac{ab}{a-b}$ and $y=y^{\prime }$

Putting the value of $x$ and $y$ in equation $\left(1\right)$, we get

$(a-b)({(x^{\prime }+\frac{ab}{a-b})}^2+y^{\prime 2})-2ab.(x^{\prime }+\frac{ab}{a-b})=0$

or, $(a-b)(x^{\prime 2}+\frac{a^2{b}^2}{(a-b{)}^2}+2\frac{x^{\prime }ab}{a-b}+y^{\prime 2})-2ab{x}^{\prime }-2\frac{a^2{b}^2}{a-b}=0$

Dividing both sides by $(a-b)$,

$(x^{\prime 2}+\frac{a^2{b}^2}{(a-b{)}^2}+2\frac{x^{\prime }ab}{a-b}+y^{\prime 2})-2\frac{x^{\prime }ab}{a-b}-2\frac{a^2{b}^2}{(a-b{)}^2}=0$

or, $x^{\prime 2}+y^{\prime 2}=\frac{a^2{b}^2}{(a-b{)}^2}$

or, $(a-b{)}^2(x^{\prime 2}+y^{\prime 2})=a^2{b}^2$