Given curves

$y = kx^2 \, ,\quad x = ky^2 \qquad (k>0)$

We need the area enclosed between them.

Find points of intersection

From $y = kx^2$ substitute into $x = ky^2$

$x = k(kx^2)^2 = k^3x^4$

So, $x = k^3x^4$ $x(1-k^3x^3)=0$

Hence, $x=0 \quad \text{or} \quad x^3=\frac1{k^3}$ $x=0 \quad \text{or} \quad x=\frac1k$

Using $y=kx^2$:

• at $x=0$, $y=0$
• at $x=\frac1k$, $y=k\left(\frac1k\right)^2=\frac1k$

So the intersection points are: $(0,0),\; \left(\frac1k,\frac1k\right)$

Write both curves as $y$ in terms of $x$

From $x=ky^2$ we get $y=\sqrt{\frac{x}{k}}$ (taking positive branch since enclosed region lies in first quadrant).

Thus the two curves are: $y = kx^2, \qquad y = \sqrt{\frac{x}{k}}$

Decide upper and lower curve

On $0 \le x \le \frac1k$, $\sqrt{\frac{x}{k}} \ge kx^2$ so area is $A = \int_0^{1/k} \left(\sqrt{\frac{x}{k}} - kx^2\right) dx$

Compute the integral

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

$A = \frac{1}{\sqrt{k}}\int_0^{1/k} x^{1/2}dx - k\int_0^{1/k} x^2dx$

Using $\int x^{1/2}dx = \frac{2}{3}x^{3/2}, \qquad \int x^2dx = \frac{x^3}{3}$

we get $A = \frac{1}{\sqrt{k}}\left[\frac{2}{3}x^{3/2}\right]_0^{1/k} - k\left[\frac{x^3}{3}\right]_0^{1/k}$

$A = \frac{1}{\sqrt{k}}\cdot \frac{2}{3}\left(\frac{1}{k}\right)^{3/2} - k\cdot \frac{1}{3}\left(\frac{1}{k}\right)^3$

$A = \frac{2}{3k^2} - \frac{1}{3k^2}$

$A = \frac{1}{3k^2}$

Use the given area

Since enclosed area is $1$ square unit, $\frac{1}{3k^2} = 1$

$3k^2 = 1$ $k^2 = \frac13$

Since $k>0$, $k = \frac{1}{\sqrt{3}}$

Match with options

$\boxed{k=\frac{1}{\sqrt{3}}}$

This is Option D.