182 If the origin $O$ is taken as one of the acute vertices to construct an isosceles right triangle inscribed in the parabola $y=a x^{2} (a>0)$ (with all three vertices on the parabola), and if the right-angle vertex $C$ is in the first quadrant, then the ratio $t\left(=\frac{y_{c}}{x_{c}}\right)$ of the ordinate to the abscissa of point $C$ must satisfy a cubic equation in $t$ which is $\qquad$ (express the cubic equation in terms of $t$).
$182 t^{3}-2 t^{2}-1=0$. Let another vertex be $A\left(x_{A}, y_{A}\right)$, then by the geometric meaning of complex numbers (where $i^{2}=-1$)
$$
\begin{aligned}
x_{A}+a \cdot x_{A}^{2} \cdot i & =\left(x_{C}+a \cdot x_{C}^{2} \cdot i\right) \cdot \sqrt{2}\left(\cos 45^{\circ}+i \cdot \sin 45^{\circ}\right) \\
& =\left(x_{C}+a \cdot x_{C}^{2} \cdot i\right) \cdot(1+i) \\
& =x_{C}-a \cdot x_{C}^{2}+\left(x_{C}+a \cdot x_{C}^{2}\right) \cdot i,
\end{aligned}
$$

By the definition of equality of complex numbers, we get
$$
\left\{\begin{array}{l}
x_{A}=x_{C}-a \cdot x_{C}^{2}, \\
a \cdot x_{A}^{2}=x_{C}+a \cdot x_{C}^{2} .
\end{array}\right.
$$

Therefore, we have
$$
a \cdot\left(x_{C}-a \cdot x_{C}^{2}\right)^{2}=x_{C}+a \cdot x_{C}^{2} \text {, }
$$

After combining both sides, we get
$$
\left(a \cdot x_{C}\right)^{3}-2 \cdot\left(a \cdot x_{C}\right)^{2}-1=0,
$$

That is, $t\left(=y_{C} / x_{C}=a \cdot x_{C}^{2} / x_{C}=a \cdot x_{C}\right)$ satisfies the cubic equation $t^{3}-2 t^{2}-1=0$.