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(C) Step $1$: Find the points of intersection of the curves $y^2 = 4kx$ and $y = kx$.Substitute $y = kx$ into $y^2 = 4kx$:$(kx)^2 = 4kx \implies k^2x^

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) Step $1$: Find the points of intersection of the curves $y^2 = 4kx$ and $y = kx$.Substitute $y = kx$ into $y^2 = 4kx$:$(kx)^2 = 4kx \implies k^2x^2 - 4kx = 0 \implies kx(kx - 4) = 0$.So, $x = 0$ or $x = \frac{4}{k}$.Step $2$: Calculate the area using integration:$	ext{Area} = \int_{0}^{4/k} (\sqrt{4kx} - kx) \, dx = \int_{0}^{4/k} (2\sqrt{k}\sqrt{x} - kx) \, dx$.Step $3$: Evaluate the integral:$	ext{Area} = [2\sqrt{k} \cdot \frac{2}{3}x^{3/2} - k \cdot \frac{x^2}{2}]_{0}^{4/k} = [\frac{4\sqrt{k}}{3}x^{3/2} - \frac{kx^2}{2}]_{0}^{4/k}$.Step $4$: Substitute the limits:$	ext{Area} = \frac{4\sqrt{k}}{3}(\frac{4}{k})^{3/2} - \frac{k}{2}(\frac{4}{k})^2 = \frac{4\sqrt{k}}{3} \cdot \frac{8}{k\sqrt{k}} - \frac{k}{2} \cdot \frac{16}{k^2} = \frac{32}{3k} - \frac{8}{k} = \frac{32 - 24}{3k} = \frac{8}{3k}$.Step $5$: Equate to the given area $\frac{2}{3}$:$\frac{8}{3k} = \frac{2}{3} \implies 8 = 2k \implies k = 4$.

Column-$I$	Column-$II$
$(A)$ In a triangle $\triangle XYZ$,let $a, b$ and $c$ be the lengths of the sides opposite to the angles $X, Y$ and $Z$,respectively. If $2(a^2-b^2)=c^2$ and $\lambda=\frac{\sin(X-Y)}{\sin Z}$,then possible values of $n$ for which $\cos(n\pi\lambda)=0$ is (are)	$(P)$ $1$
$(B)$ In a triangle $\triangle XYZ$,let $a, b$ and $c$ be the lengths of the sides opposite to the angles $X, Y$ and $Z$,respectively. If $1+\cos 2X-2\cos 2Y=2\sin X\sin Y$,then possible value$(s)$ of $\frac{a}{b}$ is (are)	$(Q)$ $2$
$(C)$ In $\mathbb{R}^2$,let $\sqrt{3}\hat{i}+\hat{j}$,$\hat{i}+\sqrt{3}\hat{j}$ and $\beta\hat{i}+(1-\beta)\hat{j}$ be the position vectors of $X, Y$ and $Z$ with respect to the origin $O$,respectively. If the distance of $Z$ from the bisector of the acute angle of $\overline{OX}$ with $\overline{OY}$ is $\frac{3}{\sqrt{2}}$,then possible value$(s)$ of $\|\beta\|$ is (are)	$(R)$ $3$
$(D)$ Suppose that $F(\alpha)$ denotes the area of the region bounded by $x=0, x=2, y^2=4x$ and $y=\|\alpha x-1\|+\|\alpha x-2\|+\alpha x$,where $\alpha \in \{0, 1\}$. Then the value$(s)$ of $F(\alpha)+\frac{8}{3}\sqrt{2}$,when $\alpha=0$ and $\alpha=1$,is (are)	$(S)$ $5$
	$(T)$ $6$

If the area enclosed between the curves $y^2 = 4kx$ and $y = kx$ for $k > 0$ is $\frac{2}{3}$ sq. units, then $k =$ ?

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