Let S denote the locus of the mid-points of t | vedclass

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(B) Let the mid-point of the chord be $(h, k)$. The equation of the chord of the parabola $y^2=x$ with mid-point $(h, k)$ is given by $T=S_1$,which is

Vedclass · Accepted Answer

$A, C$

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(B) Let the mid-point of the chord be $(h, k)$. The equation of the chord of the parabola $y^2=x$ with mid-point $(h, k)$ is given by $T=S_1$,which is $ky - \frac{1}{2}(x+h) = k^2 - h$,or $x - 2ky + 2k^2 - h = 0$.The area of the region enclosed between a parabola $y^2=4ax$ and a chord with mid-point $(h, k)$ is given by $\frac{1}{6a^2} |y_1 - y_2|^3$,where $y_1, y_2$ are the ordinates of the intersection points. For $y^2=x$,$a=\frac{1}{4}$. The area formula simplifies to $\frac{4}{3} (h-k^2)^{3/2} = \frac{4}{3}$.Thus,$(h-k^2)^{3/2} = 1$,which implies $h-k^2=1$,or $x-y^2=1$. This is the curve $S$.Checking options: For $(4, \sqrt{3})$,$4-(\sqrt{3})^2 = 4-3=1$. So $(4, \sqrt{3}) \in S$. For $(5, \sqrt{2})$,$5-(\sqrt{2})^2 = 5-2=3 
eq 1$. So $(B)$ is false.The region $R$ is bounded by $y^2=x$ and $y^2=x-1$ from $x=1$ to $x=4$.Area $= \int_1^4 (\sqrt{x} - \sqrt{x-1}) dx = [\frac{2}{3} x^{3/2} - \frac{2}{3} (x-1)^{3/2}]_1^4 = \frac{2}{3} (8 - 3\sqrt{3} - 1) = \frac{14}{3} - 2\sqrt{3}$.Thus,$(A)$ and $(C)$ are true.

List $I$	List $II$
$P. \quad m=$	$1. \quad 1/2$
$Q. \quad \text{Maximum area of } \triangle EFG \text{ is}$	$2. \quad 4$
$R. \quad y_0=$	$3. \quad 2$
$S. \quad y_1=$	$4. \quad 1$

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