jmc

Let $y=mx+c$ be a line passing through $(0,c).$ Setting $y=x^2,$ we get $$x^2=mx+c,$$or $x^2-mx-c=0.$ Let $x_1$ and $x_2$ be the roots of this equation. By Vieta's formulas, $x_1+x_2=m$ and $x_1{x}_2=-c.$

Also, $A$ and $B$ are $(x_1,m{x}_1+c)$ and $(x_2,m{x}_2+c)$ in some order, so $$\begin{array}{rl}\frac{1}{A{C}^2}+\frac{1}{B{C}^2}& =\frac{1}{x_1^2+m^2{x}_1^2}+\frac{1}{x_2^2+m^2{x}_2^2}\\ & =\frac{1}{m^2+1}\left(\frac{1}{x_1^2}+\frac{1}{x_2^2}\right)\\ & =\frac{1}{m^2+1}\cdot \frac{x_1^2+x_2^2}{x_1^2{x}_2^2}\\ & =\frac{1}{m^2+1}\cdot \frac{(x_1+x_2{)}^2-2x_1{x}_2}{(x_1{x}_2{)}^2}\\ & =\frac{1}{m^2+1}\cdot \frac{m^2+2c}{c^2}.\end{array}$$For this expression to be independent of $m,$ we must have $c=\frac{1}{2}.$ Hence, the constant $t$ is $\boxed{4}.$