The Folium of Descartes is a curve, defined by the equation $x^3 + y^3 = 3axy$, where a is a real constant. It is given that $a \ne 0$ for this question.

1. Differentiate both sides of the equation implicitly with respect to $x$.

$\frac{d}{dx}(x^3 + y^3) = \frac{d}{dx}(3axy)$
$3x^2 + 3y^2 \frac{dy}{dx} = 3a \left( y + x \frac{dy}{dx} \right)$

2. Collect the $\frac{dy}{dx}$ terms on one side and make it the subject, then simplify.

$3x^2 + 3y^2 \frac{dy}{dx} = 3ay + 3ax \frac{dy}{dx}$
$3y^2 \frac{dy}{dx} - 3ax \frac{dy}{dx} = 3ay - 3x^2$
$(3y^2 - 3ax)\frac{dy}{dx} = 3ay - 3x^2$
$\frac{dy}{dx} = \frac{3ay - 3x^2}{3y^2 - 3ax}$
$\frac{dy}{dx} = \frac{3(ay - x^2)}{3(y^2 - ax)}$
$\frac{dy}{dx} = \frac{ay - x^2}{y^2 - ax}$

1. Find the gradient of the tangent at the point $\left(\frac{3a}{2},\frac{3a}{2}\right)$.

$\frac{dy}{dx} = \frac{a\left(\frac{3a}{2}\right) - \left(\frac{3a}{2}\right)^2}{\left(\frac{3a}{2}\right)^2 - a\left(\frac{3a}{2}\right)}$\n$\frac{dy}{dx} = \frac{\frac{3a^2}{2} - \frac{9a^2}{4}}{\frac{9a^2}{4} - \frac{3a^2}{2}}$\n$\frac{dy}{dx} = \frac{-\frac{3a^2}{4}}{\frac{3a^2}{4}} = -1$

2. Find the gradient of the normal, formulate its equation, and draw the required conclusion.

Gradient of the normal, $m_{\text{normal}} = -\frac{1}{-1} = 1$\n\nEquation of the normal:\n$y - \frac{3a}{2} = 1 \left(x - \frac{3a}{2}\right)$\n$y = x$\nSince the final equation $y = x$ does not contain the constant $a$, the equation of the normal is independent of the value of $a$ (shown).

1. Use the condition for a stationary point to find a relationship between $p$ and $q$.

At a stationary point, $\frac{dy}{dx} = 0$.
$\frac{ay - x^2}{y^2 - ax} = 0 \implies ay - x^2 = 0$
Substitute $x = pa$ and $y = qa$ into this equation:
$a(qa) - (pa)^2 = 0$
$a^2q - a^2p^2 = 0$
Since $a \neq 0$, dividing by $a^2$ gives:
$q = p^2$

2. Substitute the coordinates of the stationary point into the equation of the curve.

Since the point $(pa, qa)$ lies on the curve $x^3 + y^3 = 3axy$, it must satisfy its equation:
$(pa)^3 + (qa)^3 = 3a(pa)(qa)$
$a^3p^3 + a^3q^3 = 3a^3pq$
Since $a \neq 0$, divide by $a^3$:
$p^3 + q^3 = 3pq$

3. Solve the simultaneous equations for $p$.

Substitute $q = p^2$ into $p^3 + q^3 = 3pq$:
$p^3 + (p^2)^3 = 3p(p^2)$
$p^3 + p^6 = 3p^3$
$p^6 - 2p^3 = 0$
$p^3(p^3 - 2) = 0$
Since $p$ is given as a positive constant, $p > 0$ and therefore $p \neq 0$. Thus:
$p^3 - 2 = 0 \implies p^3 = 2 \implies p = 2^{\frac{1}{3}}$

4. Find the exact value of $q$.

Substitute $p = 2^{\frac{1}{3}}$ back into $q = p^2$:
$q = \left(2^{\frac{1}{3}}\right)^2 = 2^{\frac{2}{3}}$