The Teardrop

Let the two parabolas be $y_1 = x^2 + c$ and $y_2^2 = x + c$ and the point their tangents touch be $P(a,b)$ . Then, $y_1(a) = y_2(a)$ and $\dfrac {dy_1}{dx} \bigg|_{x=a} = \dfrac {dy_2}{dx} \bigg|_{x=a}$ .

$\begin{cases} \dfrac {dy_1}{dx} = 2x & \implies \dfrac {dy_1}{dx} \bigg|_{x=a} = 2a \\ 2y_2 \dfrac {dy_2}{dx} = 1 \implies \dfrac {dy_2}{dx} = \dfrac 1{2y_2} & \implies \dfrac {dy_2}{dx} \bigg|_{x=a} = \dfrac 1{2b} \end{cases} \implies 2a = \dfrac 1{2b} \implies ab = \dfrac 14$

From symmetry, we can assume that $a=b$ , $\implies a = b = -\frac 12$ as $P$ is in the third quadrant. From $y_1(a) = b = a^2 + c$ $\implies c = - \frac 34$ .

We note that the area required is twice that bounded by either parabola and the orange straight line $y=x$ . Since it is easier to work with purple parabola $y = x^2 - \frac 34$ , we have the area:

$\begin{aligned} A & = 2 \int_{-\frac 12}^\frac 32 \left(x - x^2 + \frac 34\right) dx = 2 \left[\frac {x^2}2 - \frac {x^3}3 + \frac {3x}4 \right]_{-\frac 12}^\frac 32 = \frac 83 \end{aligned}$

$\implies A+B = 8+3 = \boxed{11}$