Question
What does the equation $(\text{a} − \text{b}) (\text{x}^2 + \text{y}^2) −2\text{abx} = 0$ become if the origin is shifted to the point $\Bigg(\frac{\text{a}\text{b}}{\text{a}-\text{b}},0\Bigg)$ without rotation?

Answer

We have,$(\text{a} − \text{b}) (\text{x}^2 + \text{y}^2) −2\text{abx} = 0$
Substituting $\text{x}=\text{x}+\frac{\text{ab}}{\text{a-b}},\text{y}=\text{Y}$ in the given equation, we get
$(\text{a}-\text{b})\Bigg[\Big(\text{x}+\frac{\text{ab}}{\text{a-b}}\Big)+\text{Y}^2\Bigg]-2\text{ab}\Bigg[\text{x}+\frac{\text{ab}}{\text{a-b}}\Bigg]=0$
$\Rightarrow (\text{a}-\text{b})\Bigg[\text{x}^2+\Big(\frac{\text{ab}}{\text{ab}}\Big)^2+2\frac{\text{xab}}{\text{a-b}}+\text{Y}^2\Bigg]-2\text{ab}\text{x}-2\frac{\text{(ab)}^2}{\text{a-b}}=0$
$\Rightarrow (\text{a-b})\Bigg[\frac{\text{x}^2(\text{a-b})^2+(\text{ab})\text{h}^2+2\text{xab}(\text{a-b})+\text{Y}^2(\text{a-b})^2}{(\text{a-b})^2}\Bigg],\frac{2\text{abx}(\text{a-b})+2(\text{ab})^2}{\text{a-b}}=0$
$\Rightarrow\frac{\text{x}^2(\text{a-b})^2+(\text{ab})^2+2\text{ab}(\text{a-b})+\text{Y}^2(\text{a-b})^2}{\text{a-b}},\frac{2\text{ab}(\text{a-b})+2(\text{ab})^2}{\text{a-b}}=0$
$\Rightarrow\text{x}^2(\text{a}-\text{b})^2+\text{Y}^2(\text{a}-\text{b})^2+(\text{ab})^2+2\text{ab}(\text{a-b})=2\text{ab}(\text{a-b})+2(\text{ab})^2$
$\Rightarrow (\text{a-b})^2(\text{x}^2+\text{Y}^2)=(\text{a}\text{b})^2$
$\Rightarrow (\text{a-b})^2(\text{x}^2+\text{Y}^2)=\text{a}^2\text{b}^2$

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