Let R be a rectangle given by the lines x=0, | vedclass

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(B) The area of the rectangle $R$ is $2 	imes 5 = 10$ square units.The line segment $AB$ cuts off a triangle $OAB$ from the rectangle,where $O$ is th

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hyperbola

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(B) The area of the rectangle $R$ is $2 	imes 5 = 10$ square units.The line segment $AB$ cuts off a triangle $OAB$ from the rectangle,where $O$ is the origin $(0,0)$.The area of triangle $OAB$ is $\frac{1}{2} 	imes \alpha 	imes \beta = \frac{\alpha \beta}{2}$.The line segment $AB$ divides the rectangle into two parts with areas in the ratio $4:1$. The area of the triangle $OAB$ is the smaller part,so its area must be $\frac{1}{5}$ of the total area of the rectangle.$\frac{	ext{Area}(OAB)}{	ext{Area}(R)} = \frac{1}{5} \implies \frac{\alpha \beta / 2}{10} = \frac{1}{5} \implies \frac{\alpha \beta}{20} = \frac{1}{5} \implies \alpha \beta = 4$.Let $M(h, k)$ be the mid-point of $AB$. Then $h = \frac{\alpha}{2}$ and $k = \frac{\beta}{2}$,which means $\alpha = 2h$ and $\beta = 2k$.Substituting these into $\alpha \beta = 4$,we get $(2h)(2k) = 4 \implies 4hk = 4 \implies hk = 1$.The locus of the mid-point $M(x, y)$ is $xy = 1$,which represents 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 $R$ be a rectangle given by the lines $x=0, x=2, y=0$ and $y=5$. Let $A(\alpha, 0)$ and $B(0, \beta)$,where $\alpha \in [0, 2]$ and $\beta \in [0, 5]$,be such that the line segment $AB$ divides the area of the rectangle $R$ in the ratio $4:1$. Then,the mid-point of $AB$ lies on a $.........$.

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