Question

Let $A (-1,0)$ and $B (2,0)$ be two points on the $x$-axis. A point ' $M$ ' is moving in $xy$-plane (other than $x$-axis) in such a way that $\angle MBA =2 \angle MAB$, then the point ' $M$ ' moves along a conic whose

• A. eccentricity equals 2
• B. vertices $( \pm 3,0)$
• C. length of latus-rectum equals 6
• D. equation of directrices are $x= \pm \frac{1}{2}$

Answer 1

Answer: (A), (C), (D)

Given, $\theta=2 \phi \Rightarrow \tan \theta=\tan 2 \phi$
$\Rightarrow \tan \theta=\frac{2 \tan \phi}{1-\tan ^2 \phi} \Rightarrow \frac{-y_0}{x_0-2}=\frac{2 y_0\left(x_0+1\right)}{\left(x_0+1\right)^2-y_0^2} \Rightarrow \frac{-1}{x_0-2}=\frac{2\left(x_0+1\right)}{\left(x_0+1\right)^2-y_0^2} $
$\Rightarrow 3 x_0{ }^2-y_0{ }^2=3$
$\therefore \text { Locus of } M \text { is hyperbola } 3 x^2-y^2=3$
$\text { Now, verify alternatives. }$