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