Parabolas and Common Tangents

From $f(x) = a_1x^2 + b_1x + c_1 \implies \begin{cases} f(0) = c_1 = 1 & & ...(0) \\ f(1) = a_1 + b_1 + c_1 = a_1 + b_1 + 1 = 3 & \implies a_1+b_1 = 2 & ...(1) \\ f(2) = 4a_1 + 2 b_1 + 1 = 9 & \implies 2a_1+b_1 = 4 & ...(2) \end{cases}$

$\implies (2)-(1): a_1 = 2, b_1 = 0 \implies f(x) = 2x^2 + 1$

Similarly, $\begin{cases} g(0) = c_2 = 3 & & ...(0) \\ g(1) = a_2 + b_2 = 1 & & ...(1) \\ g(2) = 4a_2 + 2 b_2 = 4 & \implies 2a_2+b_2 = 2 & ...(2) \end{cases}$

$\implies (2)-(1): a_2 = 2, b_2 = 0 \implies g(x) = x^2 + 3$

We note that for $B(x_B, y_B)$ , $y_B = 3$ . Then $f(x_B) = x_B^2 + 3 = 3 \implies x_B = 1$ . Therefore $B(1,3)$ and $B'(-1,3)$ . The gradient at point $B$ is $f'(x_B) = 4x_B = 4$ . Then the gradient at $A(x_A, y_A$ is also $4$ . $\implies g'(x_A) = 2x_A = 4 \implies x_A = 2$ and $y_A = g(x_A) = x_A^2 + 3 = 7$ . $\implies A(2, 7)$ and $A'(-2,7)$ .

Therefore, the area of trapezoid $A'ABB'$ is $\dfrac {A'A+B'B}2\times (y_A-Y_B) = \dfrac {2+4}2\times (7-3) = \boxed{12}$ .

$f(1) = a_{1} + b_{1} + c_{1} = 3$

$f(2) = 4a_{1} = 2b_{1} + c_{1} = 9$

$f(3) = c_{1} = 1 \implies$

$a_{1} + b_{1} = 2$

$2a_{1} + b_{1} = 4$

$\implies a_{1} = 2$ and $b_{1} = 0$

$\implies \boxed{f(x) = 2x^2 + 1}$ .

$g(1) = a_{2} b_{2} + c{2} = 4$

$g(2) = 4a_{2} + 2b_{2} + c-{2} = 7$

$g(0) = c_{2} = 3 \implies$

$a_{2} + b_{2} = 1$

$2a_{2} + b_{2} = 2$

$\implies a_{2} = 1$ and $b_{2} = 0$

$\implies \boxed{g(x) = x^2 + 3}$

$f'(x_{0}) = 4x_{0} = 2x_{1} = g'(x_{1}) \implies x_{1} = 2x_{0}$

$B : (x_{0},2x_{0}^2 + 1)$ and $A : (2x_{0},4x_{0}^2 + 3)$

$\implies m_{AB} = \dfrac{2x_{0}^2 + 2}{x_{0}} = 4x_{0} \implies 4x_{0}^2 = 2x-{0}^2 + 2 \implies$

$x_{0} = \pm 1$

$x_{0} = 1 \implies B : (1,3)$ and $A : (2,7)$

and

$x_{0} = -1 \implies B' : (-1,3)$ and $A' : (-2,7)$

$\implies A'A = 4, \:\ B'B = 2$ and the height of $h$ of trapezoid $A'ABB'$ is $h = 4$

$\implies$ The area of trapezoid $A'ABB'$ is $A_{A'ABB'} = \dfrac{1}{2}(4)(2 + 4) = \boxed{12}$ .