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Two points $A(-a, 0)$ and $B(a, 0)$ are given. If $C$ is a variable point lying on one side of the line $A B$ such that $\angle C A B-\angle C B A=\alpha$, where $\alpha$ is a positive constant, then locus of the point $C$ is
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Verified Answer
The correct answer is:
$a^2-x^2+y^2+2 x y \cot \alpha=0$
Given
Let $C(x, y)$
Slope of $A B=0$
$$
\begin{aligned}
\therefore \tan \beta & =\text { Slope of } A C \\
\therefore \tan \beta & =\frac{y}{x+a}
\end{aligned}
$$
Slope of $B C=\tan (\pi-\gamma)$
$$
\begin{aligned}
& \Rightarrow \quad-\tan \gamma=\frac{y}{x-a} \\
& \Rightarrow \quad \tan \gamma=\frac{-y}{x-a}
\end{aligned}
$$
Given,
$$
\begin{array}{ll}
\because \quad \tan (\beta-\gamma)=\frac{\tan \beta-\tan \gamma}{1+\tan \beta \tan \gamma} \\
\Rightarrow \quad \tan \alpha=\frac{\frac{y}{x+a}+\frac{y}{x-a}}{1-\frac{y^2}{x^2-a^2}} \\
\Rightarrow & \tan \alpha=\frac{y(x-a+x+a)}{x^2-a^2-y^2}
\end{array}
$$
$$
\beta-\gamma=\alpha
$$
$$
\begin{array}{lr}
\Rightarrow & \tan \alpha=\frac{2 x y}{x^2-a^2-y^2} \\
\Rightarrow & x^2-y^2-2 x y \cot \alpha-a^2=0 \\
\Rightarrow & a^2-x^2+y^2+2 x y \cot \alpha=0
\end{array}
$$
Let $C(x, y)$
Slope of $A B=0$
$$
\begin{aligned}
\therefore \tan \beta & =\text { Slope of } A C \\
\therefore \tan \beta & =\frac{y}{x+a}
\end{aligned}
$$
Slope of $B C=\tan (\pi-\gamma)$
$$
\begin{aligned}
& \Rightarrow \quad-\tan \gamma=\frac{y}{x-a} \\
& \Rightarrow \quad \tan \gamma=\frac{-y}{x-a}
\end{aligned}
$$
Given,
$$
\begin{array}{ll}
\because \quad \tan (\beta-\gamma)=\frac{\tan \beta-\tan \gamma}{1+\tan \beta \tan \gamma} \\
\Rightarrow \quad \tan \alpha=\frac{\frac{y}{x+a}+\frac{y}{x-a}}{1-\frac{y^2}{x^2-a^2}} \\
\Rightarrow & \tan \alpha=\frac{y(x-a+x+a)}{x^2-a^2-y^2}
\end{array}
$$
$$
\beta-\gamma=\alpha
$$
$$
\begin{array}{lr}
\Rightarrow & \tan \alpha=\frac{2 x y}{x^2-a^2-y^2} \\
\Rightarrow & x^2-y^2-2 x y \cot \alpha-a^2=0 \\
\Rightarrow & a^2-x^2+y^2+2 x y \cot \alpha=0
\end{array}
$$
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