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

• A

  a straight line
• B

  a parabola
• C

  an ellipse
• Correct answer
  D

  a hyperbola

Solution

$$ \begin{array}{l} \text { Let } \angle M A B=\theta \text { , then } \angle M B A=2 \theta \\ \text { tan } \theta=\frac{k}{1+h} \text { and } \tan 2 \theta=\frac{k}{2-h} \\ \text { then } \tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^{2} \theta}=\frac{k}{2-h} \\ \Rightarrow \quad \frac{2(k / 1+h)}{1-(k / 1+h)^{2}}=\frac{k}{2-h} \\ \Rightarrow \quad \frac{2}{1+h} \times \frac{(1+h)^{2}}{(1+h)^{2}-k^{2}}=\frac{1}{2-h} \\ \Rightarrow \quad \frac{2(1+h)}{(1+h)^{2}-k^{2}}=\frac{1}{2-h} \\ \Rightarrow \quad 1+h^{2}+2 h-k^{2}=2(1+h)(2-h) \\ \Rightarrow \quad 1+h^{2}+2 h-k^{2}=2\left(2-h+2 h-h^{2}\right) \\ \Rightarrow \quad 1+h^{2}+2 h-k^{2}=2\left(2+h-h^{2}\right) \\ \Rightarrow \quad 1+h^{2}+2 h-k^{2}=4+2 h-2 h^{2} \\ \Rightarrow \quad 1+3 h^{2}-k^{2}=4 \\ \Rightarrow \quad 3 h^{2}-k^{2}=3 \end{array} $$ which represents hyperbola (d).

Hence, the correct answer is option: 'D'.