If the area of the region bounded by the curves $y=x^2$ and $x=y^2$ is $k$, then the area of the region bounded by the curves $\frac{x+\sqrt{3} y}{2}=\left(\frac{\sqrt{3} x-y}{2}\right)^2$ and $\frac{\sqrt{3} x-y}{2}=\left(\frac{x+\sqrt{3} y}{2}\right)^2$, is

Question

If the area of the region bounded by the curves $y=x^2$ and $x=y^2$ is $k$, then the area of the region bounded by the curves $\frac{x+\sqrt{3} y}{2}=\left(\frac{\sqrt{3} x-y}{2}\right)^2$ and $\frac{\sqrt{3} x-y}{2}=\left(\frac{x+\sqrt{3} y}{2}\right)^2$, is, $\frac{\sqrt{3}}{2} k$, $\frac{1}{2} k$, $k$, $\left(\frac{\sqrt{3}+1}{2}\right) k$

MARKS App · Accepted Answer

Let $\frac{x+\sqrt{3} y}{2}=\frac{x+\sqrt{3} y}{\sqrt{1+(\sqrt{3})^2}}=y$ and $\frac{\sqrt{3} x-y}{2}=\frac{\sqrt{3} x-y}{\sqrt{(\sqrt{3})^2+1}}=x$ so, the given curve $\frac{x+\sqrt{3} y}{2}=\left(\frac{\sqrt{3} x-y}{2}\right)^2$, becomes $y=x^2$...(i) Similarly, the given curve $\frac{\sqrt{3} x-y}{2}=\left(\frac{x+\sqrt{3} y}{2}\right)^2$, becomes $x^2=y$...(ii) So the area of the region bounded by curves (i) and (ii) is equals to the area of the region bounded by the curve $y=x^2$ and $x=y^2$ is equal to ‘ $K$ ' (given).

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