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Let $A(-1,0)$ and $B(2,0)$ be two points. $A$ point $M$ moves in the plane in such a way that $\angle M B A=2 \angle M A B .$ Then, the point $M$ moves along
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The correct answer is:
a hyperbola
$$
\begin{array}{l}
\text { Let } \angle M A B=\theta \text { , then } \angle M B A=2 \theta \\
\text { tan } \theta=\frac{k}{1+h} \text { and } \tan 2 \theta=\frac{k}{2-h} \\
\text { then } \tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^{2} \theta}=\frac{k}{2-h} \\
\Rightarrow \quad \frac{2(k / 1+h)}{1-(k / 1+h)^{2}}=\frac{k}{2-h} \\
\Rightarrow \quad \frac{2}{1+h} \times \frac{(1+h)^{2}}{(1+h)^{2}-k^{2}}=\frac{1}{2-h} \\
\Rightarrow \quad \frac{2(1+h)}{(1+h)^{2}-k^{2}}=\frac{1}{2-h} \\
\Rightarrow \quad 1+h^{2}+2 h-k^{2}=2(1+h)(2-h) \\
\Rightarrow \quad 1+h^{2}+2 h-k^{2}=2\left(2-h+2 h-h^{2}\right) \\
\Rightarrow \quad 1+h^{2}+2 h-k^{2}=2\left(2+h-h^{2}\right) \\
\Rightarrow \quad 1+h^{2}+2 h-k^{2}=4+2 h-2 h^{2} \\
\Rightarrow \quad 1+3 h^{2}-k^{2}=4 \\
\Rightarrow \quad 3 h^{2}-k^{2}=3
\end{array}
$$
which represents hyperbola (d).
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