Two points A(-a, 0) and B(a, 0) are given. If | vedclass

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(B) Let $C(x, y)$ be the variable point. Let $\angle CAB = \beta$ and $\angle CBA = \gamma$.The slope of $AC$ is $m_1 = \frac{y-0}{x-(-a)} = \frac{y}{

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$a^2-x^2+y^2+2xy \cot \alpha=0$

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(B) Let $C(x, y)$ be the variable point. Let $\angle CAB = \beta$ and $\angle CBA = \gamma$.The slope of $AC$ is $m_1 = \frac{y-0}{x-(-a)} = \frac{y}{x+a} = 	an \beta$.The slope of $BC$ is $m_2 = \frac{y-0}{x-a} = \frac{y}{x-a}$. Since $\angle CBA = \gamma$,the slope of $BC$ is $-	an \gamma$ (as it makes an obtuse angle with the positive $x$-axis).Thus,$	an \gamma = -\frac{y}{x-a} = \frac{y}{a-x}$.Given $\beta - \gamma = \alpha$,we have $	an(\beta - \gamma) = 	an \alpha$.Using the formula $	an(\beta - \gamma) = \frac{	an \beta - 	an \gamma}{1 + 	an \beta 	an \gamma}$:$	an \alpha = \frac{\frac{y}{x+a} - \frac{y}{a-x}}{1 + (\frac{y}{x+a})(\frac{y}{a-x})} = \frac{\frac{y(a-x) - y(x+a)}{(x+a)(a-x)}}{1 + \frac{y^2}{a^2-x^2}} = \frac{\frac{ay - xy - xy - ay}{a^2-x^2}}{\frac{a^2-x^2+y^2}{a^2-x^2}} = \frac{-2xy}{a^2-x^2+y^2}$.Therefore,$	an \alpha = \frac{-2xy}{a^2-x^2+y^2}$.Rearranging gives: $(a^2-x^2+y^2) 	an \alpha = -2xy$.Dividing by $	an \alpha$ (or multiplying by $\cot \alpha$): $a^2-x^2+y^2 = -2xy \cot \alpha$.Thus,$a^2-x^2+y^2+2xy \cot \alpha = 0$.

Two points $A(-a, 0)$ and $B(a, 0)$ are given. If $C$ is a variable point lying on one side of the line $AB$ such that $\angle CAB - \angle CBA = \alpha$,where $\alpha$ is a positive constant,then the locus of the point $C$ is

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