To get a handle on the constant $t,$ we can look at some particular cases.

Suppose we let $AB$ approach a vertical line.  Then $\frac{1}{AC}$ approaches 0, and $B$ approaches $(0,0),$ so $\frac{1}{AC} + \frac{1}{BC}$ approaches $c.$  Hence,
\[t = \frac{1}{c}.\]Now, suppose we take $A = (\sqrt{c},c)$ and $B = (-\sqrt{c},c).$  Then
\[t = \frac{1}{AC} + \frac{1}{BC} = \frac{1}{\sqrt{c}} + \frac{1}{\sqrt{c}} = \frac{2}{\sqrt{c}}.\]Hence, $\frac{1}{c} = \frac{2}{\sqrt{c}},$ so $\sqrt{c} = \frac{1}{2},$ and $c = \frac{1}{4}.$  Therefore, $t = 4.$  (Note that this makes $C$ the focus of the parabola.)

For a full solution, let's check that this value works.  Let $y = mx + \frac{1}{4}$ be the equation of line $AB.$  Setting $y = x^2,$ we get
\[x^2 = mx + \frac{1}{4},\]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 $x_1 x_2 = -\frac{1}{4}.$

Also, $A$ and $B$ are $(x_1,x_1^2)$ and $(x_2,x_2^2)$ in some order, so
\begin{align*}
\frac{1}{AC} + \frac{1}{BC} &= \frac{1}{\sqrt{x_1^2 + (x_1^2 - \frac{1}{4})^2}} + \frac{1}{\sqrt{x_2^2 + (x_2^2 - \frac{1}{4})^2}} \\
&= \frac{1}{\sqrt{x_1^2 + x_1^4 - \frac{1}{2} x_1^2 + \frac{1}{16}}} + \frac{1}{\sqrt{x_2^2 + x_2^4 - \frac{1}{2} x_2^2 + \frac{1}{16}}} \\
&= \frac{1}{\sqrt{x_1^4 + \frac{1}{2} x_1^2 + \frac{1}{16}}} + \frac{1}{\sqrt{x_2^4 + \frac{1}{2} x_2^2 + \frac{1}{16}}} \\
&= \frac{1}{\sqrt{(x_1^2 + \frac{1}{4})^2}} + \frac{1}{\sqrt{(x_2^2 + \frac{1}{4})^2}} \\
&= \frac{1}{x_1^2 + \frac{1}{4}} + \frac{1}{x_2^2 + \frac{1}{4}}.
\end{align*}We have that $x_1^2 x_2^2 = (x_1 x_2)^2 = \left( -\frac{1}{4} \right)^2 = \frac{1}{16}$ and
\[x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2 = m^2 + \frac{1}{2}.\]Hence,
\begin{align*}
\frac{1}{x_1^2 + \frac{1}{4}} + \frac{1}{x_2^2 + \frac{1}{4}} &= \frac{x_1^2 + \frac{1}{4} + x_2^2 + \frac{1}{4}}{(x_1^2 + \frac{1}{4})(x_2^2 + \frac{1}{4})} \\
&= \frac{x_1^2 + x_2^2 + \frac{1}{2}}{x_1^2 x_2^2 + \frac{1}{4} (x_1^2 + x_2^2) + \frac{1}{16}} \\
&= \frac{m^2 + 1}{\frac{1}{16} + \frac{1}{4} (m^2 + \frac{1}{2}) + \frac{1}{16}} \\
&= \frac{m^2 + 1}{\frac{1}{4} m^2 + \frac{1}{4}} \\
&= 4.
\end{align*}
상수 $t,$를 다루기 위해 몇 가지 특정 경우를 살펴볼 수 있습니다.

AB$가 수직선에 가까워진다고 가정해봅시다. 그러면 $\frac{1}{AC}$는 0에 가까워지고, $B$는 $(0,0)에 가까워지므로 $\frac{1}{AC} + \frac{1}{BC}$ 는 $c.$ 에 접근합니다,
\[t = \frac{1}{c}.\]이제 $A = (\sqrt{c},c)$ 와 $B = (-\sqrt{c},c)$ 를 취한다고 가정하면 다음과 같습니다.
\[t = \frac{1}{AC} + \frac{1}{BC} = \frac{1}{\sqrt{c}} + \frac{1}{\sqrt{c}} = \frac{2}{\sqrt{c}}.\]따라서 $\frac{1}{c} = \frac{2}{\sqrt{c}},$ 따라서 $\sqrt{c} = \frac{1}{2},$ 그리고 $c = \frac{1}{4}입니다.따라서 $t = 4.$입니다(이렇게 하면 $C$가 포물선의 초점이 됩니다.).

완전한 솔루션을 위해 이 값이 제대로 작동하는지 확인해 보겠습니다. y = mx + \frac{1}{4}$ 를 선 $AB.$ 의 방정식으로 가정하고 $y = x^2,$ 를 설정하면 다음과 같이 됩니다.
\x^2 = mx + \frac{1}{4},\]또는 $x^2 - mx - c = 0.$ 이 방정식의 근은 $x_1$과 $x_2$가 됩니다.  비에타의 공식에 따르면 $x_1 + x_2 = m$ 및 $x_1 x_2 = -\frac{1}{4}.$입니다.

또한 $A$와 $B$는 어떤 순서로 $(x_1,x_1^2)$와 $(x_2,x_2^2)$이므로 다음과 같습니다.
\begin{align*}
\frac{1}{AC} + \frac{1}{BC} &= \frac{1}{\sqrt{x_1^2 + (x_1^2 - \frac{1}{4})^2}} + \frac{1}{\sqrt{x_2^2 + (x_2^2 - \frac{1}{4})^2}} \\
&= \frac{1}{\sqrt{x_1^2 + x_1^4 - \frac{1}{2} x_1^2 + \frac{1}{16}}} + \frac{1}{\sqrt{x_2^2 + x_2^4 - \frac{1}{2} x_2^2 + \frac{1}{16}}} \\
&= \frac{1}{\sqrt{x_1^4 + \frac{1}{2} x_1^2 + \frac{1}{16}}} + \frac{1}{\sqrt{x_2^4 + \frac{1}{2} x_2^2 + \frac{1}{16}}} \\
&= \frac{1}{\sqrt{(x_1^2 + \frac{1}{4})^2}} + \frac{1}{\sqrt{(x_2^2 + \frac{1}{4})^2}} \\
&= \frac{1}{x_1^2 + \frac{1}{4}} + \frac{1}{x_2^2 + \frac{1}{4}}.
\end{align*}$x_1^2 x_2^2 = (x_1 x_2)^2 = \left( -\frac{1}{4} \right)^2 = \frac{1}{16}$이고, 다음과 같습니다.
\[x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2 = m^2 + \frac{1}{2}.\]따라서,
\begin{align*}
\frac{1}{x_1^2 + \frac{1}{4}} + \frac{1}{x_2^2 + \frac{1}{4}} &= \frac{x_1^2 + \frac{1}{4} + x_2^2 + \frac{1}{4}}{(x_1^2 + \frac{1}{4})(x_2^2 + \frac{1}{4})} \\
&= \frac{x_1^2 + x_2^2 + \frac{1}{2}}{x_1^2 x_2^2 + \frac{1}{4} (x_1^2 + x_2^2) + \frac{1}{16}} \\
&= \frac{m^2 + 1}{\frac{1}{16} + \frac{1}{4} (m^2 + \frac{1}{2}) + \frac{1}{16}} \\
&= \frac{m^2 + 1}{\frac{1}{4} m^2 + \frac{1}{4}} \\
&= 4.
\end{align*}