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