If the area enclosed between the curves y=kx^2 and x=ky^2, (k0), is 1 square unit, then k…

Question

If the area enclosed between the curves $y=kx^2$ and $x=ky^2$, ($k>0$), is $1$ square unit, then $k$ is:

Accepted Answer

The curves meet at $O(0,0)$ and $\left(\dfrac1k,\dfrac1k\right)$.

$$1 = \int_0^{1/k}\left(\sqrt{\frac{x}{k}}-kx^2\right)dx = \frac{1}{\sqrt k}\cdot\frac{2}{3}\left(\frac1k\right)^{3/2} - \frac{k}{3}\left(\frac1k\right)^3 = \frac{2}{3k^2}-\frac{1}{3k^2} = \frac{1}{3k^2}$$

$$3k^2 = 1 \implies k = \frac{1}{\sqrt3} \quad (k>0)$$

Solutions · 1