Express the equation $\;$ $r^{\frac{1}{2}} = a^{\frac{1}{2}} \cos \left(\dfrac{\theta}{2}\right)$ $\;$ in Cartesian coordinates.
The relation between Cartesian coordinates $\left(x, y\right)$ and polar coordinates $\left(r, \theta\right)$ is
$x = r \cos \theta$, $\;\;$ $y = r \sin \theta$
$\implies$ $\cos \theta = \dfrac{x}{r}$, $\;\;$ $\sin \theta = \dfrac{y}{r}$
Now, $\;$ $\cos \theta = 2 \cos^2 \left(\dfrac{\theta}{2}\right) - 1$
$\therefore \;$ $\cos \left(\dfrac{\theta}{2}\right) = \sqrt{\dfrac{1 + \cos \theta}{2}}$
Given equation is $\;\;$ $r^{\frac{1}{2}} = a^{\frac{1}{2}} \cos \left(\dfrac{\theta}{2}\right)$
i.e. $\;$ $\sqrt{r} = \sqrt{a} \times \sqrt{\dfrac{1 + \cos \theta}{2}}$
i.e. $\;$ $\sqrt{r} = \sqrt{a} \times \sqrt{\dfrac{1 + \frac{x}{r}}{2}}$
i.e. $\;$ $\sqrt{r} = \sqrt{a} \times \sqrt{\dfrac{x + r}{2r}}$
i.e. $\;$ $r = \sqrt{a} \times \sqrt{\dfrac{x + r}{2}}$
i.e. $\;$ $2 r^2 = a \left(x + r\right)$ $\;\;\; \cdots \; (1)$
Now, $\;$ $r^2 = x^2 + y^2$ $\;\;\; \cdots \; (2a)$ $\implies$ $r = \sqrt{x^2 + y^2}$ $\;\;\; \cdots \; (2b)$
In view of equations $(2a)$ and $(2b)$ equation $(1)$ becomes
$2 \left(x^2 + y^2\right) = a \left(x + \sqrt{x^2 + y^2}\right)$
i.e. $\;$ $2 \left(x^2 + y^2\right) - ax = a \sqrt{x^2 + y^2}$
i.e. $\;$ $\left[2 \left(x^2 + y^2\right) - ax\right]^2 = a^2 \left(x^2 + y^2\right)$
$\therefore \;$ The given equation in Cartesian coordinates is $\;$ $\left[2 \left(x^2 + y^2\right) - ax\right]^2 = a^2 \left(x^2 + y^2\right)$