It's an integer right?

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Consider a symmetric case. So T1T2 meet axis perpendicularly at M. Tangents meet at B on the axis. And tangent AC perpendicular to axis at vertex O.

Property of parabola: MO = OB.

Thus T2M = 2 OC = 2 OA.

Hence AT2 intersects OM in ratio 1:2, giving FM = 2 OF = 2a.

This gives x coordinate of T1, T2 as 3a. So the y coordinates of T1, T2 will be $+\sqrt{12}a$ and $-\sqrt{12}a$

Slope of tangent to $y^2 = 4ax$ will be $y' = \frac{2a}{y} = \frac{2a}{\sqrt{12}a} = \frac{1}{\sqrt{3}}$

Hence inclination of tangent BA = 30 degrees! Giving the desired angle to be 60 degrees. And the triangle ABC to be an equilateral triangle.

Angle $= 60° = \pi/3$ giving $\alpha = 3$ Hence $\alpha ^ \alpha - 8 = 3^3 - 8 = 27 - 8 = 19$

$\triangle ABC$ with sides tangential to the parabola $y^2=4ax$ has to be an isosceles triangle as shown above. $AC$ is along the y-axis and $B$ on the x-axis.

Let $y_D$ of $D (x_D, y_D)$ be $b$ , then $x_D = \dfrac {b^2}{4a}\space \Rightarrow D \left(\frac {b^2}{4a}, b \right)$ .

The gradient of the parabola is given by:

$2y \dfrac {dy}{dx} = 4a\quad \Rightarrow \dfrac {dy}{dx} = \dfrac {2a}{y}$

At point $D$ , gradient is $m = \dfrac {2a}{b}$ .

Therefore, for point $B$ , we note that:

$\dfrac {b}{x_D-x_B} = m = \dfrac {2a}{b}$

$\Rightarrow x_B = x_D - \dfrac {b^2}{2a} = \dfrac {b^2}{4a} - \dfrac {b^2}{2a} = - \dfrac {b^2}{4a} = - x_D$

We note that point $C$ is mid-point of $DB$ and therefore, $C \left( 0, \frac{b}{2} \right)$ and $A \left(0, -\frac{b}{2} \right)$ .

Considering the line $AD$ passing throught the focus $(a,0)$ , we have:

$\dfrac {a}{\frac{b}{2}} = \dfrac {\frac {b^2}{4a}-a}{b}\quad \Rightarrow 2a = \frac {b^2}{4a}-a\quad \Rightarrow b^2 = 12a^2 \quad \Rightarrow b = 2\sqrt{3}a$

Let $\angle ABC = \theta$ , then $\tan {\frac{\theta}{2}} = \dfrac {b}{\frac{b^2}{2a}} = \dfrac {2a}{b} = \dfrac {2a}{2\sqrt{3}a} = \dfrac {1}{\sqrt 3}$

$\Rightarrow \angle ABC = \theta = 2 \tan ^{-1} {\left( \frac {1}{\sqrt{3}}\right)} = 2\times \dfrac {\pi}{6} = \dfrac {\pi}{3}$

$\Rightarrow \alpha = 3 \quad \Rightarrow \alpha^\alpha - 8 = \boxed{19}$ .