I took B as $B(x, x^2)$. So far, I tried using the distance formula to get an expression for $|AB|^2 + |BC|^2$. I got $(3 - x)^2 + (1 - x^2)^2 + (x - 4)^2 + (x^2 - 3)^2$. Expanding this out gives $2x^4 - 6x^2 - 14x + 35$. I defined a function $f(x) = 2x^4 - 6x^2 - 14x + 35$. Then, $f'(x) = 8x^3 - 12x - 14$. To get the minimum point of $f(x)$, I tried finding the root of $8x^3 - 12x - 14 = 0$.
$4x^3 - 6x - 7 = 0$, which gives $x = 1.609$ to 4.s.f.
$2(1.609)^4 - 6(1.609)^2 - 14(1.609) + 35 = 10.345324$
I was wondering if there is a way to get an exact answer to this question instead of a decimal.
There is a way to solve this problem without calculus. We can make a system of equations and equate coefficients. We have a family of curves given by
$$ E_{\lambda}(x,y)=(x-x_A)^2+(y-y_A)^2+(x-x_B)^2+(y-y_B)^2=\lambda $$
and we need to find $\lambda = \lambda^*$ such that $E_{\lambda^*}(x,y)$ and $y-x^2=0$ are tangent. Eliminating $y$ from
$$ \cases{ (x-x_A)^2+(y-y_A)^2+(x-x_B)^2+(y-y_B)^2=\lambda\\ y=x^2 } $$
we obtain after substitution of $x_A,y_A,x_B,y_B$
$$ 2 x^4-6 x^2-14 x+35-\lambda = 0 $$
At tangency we have
$$ 2 x^4-6 x^2-14 x+35-\lambda=(x-r)^2(c_0x^2+c_1 x + c_2) $$
See the double root $r$.
$$ 2 x^4-6 x^2-14 x+35-\lambda-(x-r)^2(c_0x^2+c_1 x + c_2)=0 \ \ \forall\ \ x $$
so we need
$$ \cases{ c_2 r^2+\lambda -35=0\\ c_1 r^2-2 c_2 r+14=0\\ c_0 r^2-2 c_1 r+c_2+6=0\\ 2 c_0r-c_1=0\\ 2-c_0=0 } $$
Now solving for $\lambda, r, c_0,c_1,c_2$ we obtain $\lambda^*=10.3453$. The degree of the involved polynomial (in $r$) entails a numerical solution.