Let A(-1, 0) and B(2, 0) be two points. A poi | vedclass

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(D) Let the coordinates of point $M$ be $(h, k)$. Let $\angle MAB = 	heta$, then $\angle MBA = 2	heta$.From the coordinates of $A(-1, 0)$ and $B(2,

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a hyperbola

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(D) Let the coordinates of point $M$ be $(h, k)$. Let $\angle MAB = 	heta$, then $\angle MBA = 2	heta$.From the coordinates of $A(-1, 0)$ and $B(2, 0)$, we have:$	an 	heta = \frac{k}{h - (-1)} = \frac{k}{h+1}$$	an(\pi - 2	heta) = \frac{k}{h-2} \implies -	an 2	heta = \frac{k}{h-2} \implies 	an 2	heta = \frac{k}{2-h}$Using the identity $	an 2	heta = \frac{2 	an 	heta}{1 - 	an^2 	heta}$, we get:$\frac{k}{2-h} = \frac{2(\frac{k}{h+1})}{1 - (\frac{k}{h+1})^2}$$\frac{k}{2-h} = \frac{2k(h+1)}{(h+1)^2 - k^2}$Assuming $k 
eq 0$, we divide by $k$:$\frac{1}{2-h} = \frac{2(h+1)}{(h+1)^2 - k^2}$$(h+1)^2 - k^2 = 2(h+1)(2-h)$$h^2 + 2h + 1 - k^2 = 2(2h - h^2 + 2 - h) = 2(-h^2 + h + 2) = -2h^2 + 2h + 4$$3h^2 - k^2 = 3$This is the equation of a hyperbola.

List-$I$	List-$II$
$P$. The length of the conjugate axis of $H$ is	$1$. $8$
$Q$. The eccentricity of $H$ is	$2$. $\frac{4}{\sqrt{3}}$
$R$. The distance between the foci of $H$ is	$3$. $\frac{2}{\sqrt{3}}$
$S$. The length of the latus rectum of $H$ is	$4$. $4$

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

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