Question

Let $A(-1,0)$ and $B(2,0)$ be two points. A point $M$ moves in the plane in such a way that $\angle MBA=2\angle MAB$. Then the point $M$ moves along

• A. A straight line
• B. A parabola
• C. An ellipse
• D. A hyperbola

Answer 1

The correct option is D A hyperbola

We know that $\angle MBA=2\angle MAB$
Also $\tan \theta =\left|\frac{m_2-m_1}{1+m_1\cdot m_2}\right|$ where $m_1$ and $m_2$ are the slopes of two lines and $\theta$ is the angle between the two lines.
Let the coordinates of M be $(h,k)$
Slope of AM $=\frac{k-0}{h+1}=\frac{k}{h+1}$
Slope of AB $=\frac{0-0}{2+1}=0$
Slope of BM $=\frac{k-0}{h-2}=\frac{k}{h-2}$
$\tan \left(\angle MAB\right)=\left|\frac{0-\frac{k}{h+1}}{1}\right|=\pm \frac{k}{h+1}$

$\tan \left(\angle MBA\right)=\left|\frac{0-\frac{k}{h-2}}{1}\right|=\pm \frac{k}{h-2}$
Since $\angle MBA=2\angle MAB$
And we know that $\tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }$

If $\tan \left(\angle MAB\right)=\frac{k}{h+1}$ and $\tan \left(\angle MBA\right)=\frac{k}{h-2}$
Therefore $\frac{k}{h-2}=\frac{2k(h+1)}{(h+1{)}^2-k^2}$
$(h+1{)}^2-k^2=2(h+1\left)\right(h-2)$
$h^2+2h+1-k^2=2h^2-2h-4$
$h^2-4h-5+k^2=0$
$(h-2{)}^2+k^2=9$
The above equation represents a circle with center $(2,0)$ and radius $3$.

Which is not possible as $M$ will be fixed. So, it does not move.

If $\tan \left(\angle MAB\right)=-\frac{k}{h+1}$ and $\tan \left(\angle MBA\right)=\frac{k}{h-2}$

Or

If $\tan \left(\angle MAB\right)=\frac{k}{h+1}$ and $\tan \left(\angle MBA\right)=-\frac{k}{h-2}$

Therefore $-\frac{k}{h-2}=\frac{2k(h+1)}{(h+1{)}^2-k^2}$
$-\left\{\right(h+1{)}^2-k^2\}=2(h+1\left)\right(h-2)$
$-h^2-2h-1+k^2=2h^2-2h-4$
$3h^2-k^2=3$
$\frac{h^2}{1}-\frac{k^2}{3}=1$

This represents a hyperbola.