Simple Question Complicated Answer

The given points $(2, 2\sqrt{3})$ and $(4, 0)$ and the intersection of the two tangent lines $(0, 0)$ are points of an equilateral triangle with sides of $4$ . Therefore, the whole system can be rotated counterclockwise $60°$ so that the parabola has a vertical axis, and then shifted so that it has a vertex at the origin, for an equation of $y = ax^2$ , while still preserving its distance between the vertex and its focus.

Since the equilateral triangle had sides of $4$ , the new parabola $y = ax^2$ will have a slope of $m = \sqrt{3}$ at $x = 2$ . Since its slope is $m = 2ax$ , $\sqrt{3} = 2 \cdot a \cdot 2$ , which solve to $a = \frac{\sqrt{3}}{4}$ , for an equation of $y = \frac{\sqrt{3}}{4}x^2$ . Therefore, the distance between the vertex and focus is $p = \frac{1}{4 \cdot \frac{\sqrt{3}}{4}} = \frac{1}{\sqrt{3}} \approx \boxed{0.5773}$ .

Let (h, k) be the vertex of the parabola. Since the lines y=0 and y=x√3 are the tangents to it, it's axis has the slope $\dfrac{1}{√3}$ . Therefore it's equation is [√3(y-k)-(x-h) ]^2=8a[√3(x-h)+(y-k)], where a is the required distance. Since the line y=0 is the tangent to the curve at (4,0), we have 4√3a=4, or a= $\dfrac{1}{√3}$ =0.57735...

Let $\vec{P}_0$ be the vertex location, and let $\vec{u}_1$ and $\vec{u}_2$ be orthogonal unit vectors which effectively define a new coordinate system. Define tangent points $(x_1,y_1) = (2,2 \sqrt{3})$ and $(x_2,y_2) = (4,0)$ . The coordinates of a point on the parabola are:

$\vec{P} = \vec{P}_0 + \alpha \, \vec{u}_1 + a \, \alpha^2 \, \vec{u}_2 \\ x = x_0 + \alpha \, \vec{u}_{1x} + a \, \alpha^2 \, \vec{u}_{2x} \\ y = y_0 + \alpha \, \vec{u}_{1y} + a \, \alpha^2 \, \vec{u}_{2y} \\ \vec{u}_1 = ( \cos \theta, \sin \theta) \\ \vec{u}_2 = ( \sin \theta, -\cos \theta)$

Each point of tangency has an associated $\alpha$ value. So the six unknowns are $(x_0, y_0, \theta, a, \alpha_1, \alpha_2)$ . We need to solve six equations for the six unknowns. The first four have to do with the coordinates of the two tangency points.

$x_1 = x_0 + \alpha_1 \, \vec{u}_{1x} + a \, \alpha_1^2 \, \vec{u}_{2x} \\ y_1 = y_0 + \alpha_1 \, \vec{u}_{1y} + a \, \alpha_1^2 \, \vec{u}_{2y} \\ x_2 = x_0 + \alpha_2 \, \vec{u}_{1x} + a \, \alpha_2^2 \, \vec{u}_{2x} \\ y_2 = y_0 + \alpha_2 \, \vec{u}_{1y} + a \, \alpha_2^2 \, \vec{u}_{2y}$

Consider the tangential intersection with the $x$ axis. For $y = 0$ , there should only be one corresponding $\alpha$ value. The quadratic equation for $\alpha$ (given $y = 0$ ) should only have one solution. This gives us the fifth equation.

$0 = y_0 + \alpha \, \vec{u}_{1y} + a \, \alpha^2 \, \vec{u}_{2y} \\ A = a \, \vec{u}_{2y} \\ B = \vec{u}_{1y} \\ C = y_0 \\ B^2 - 4 A C = 0$

We can do the same thing for the tangential intersection with the line $y = \sqrt{3} x$ , yielding the sixth equation.

$y = \sqrt{3} x \\ y_0 + \alpha \, \vec{u}_{1y} + a \, \alpha^2 \, \vec{u}_{2y} = \sqrt{3} \, x_0 + \sqrt{3} \, \alpha \, \vec{u}_{1x} + \sqrt{3} \, a \, \alpha^2 \, \vec{u}_{2x} \\ A' = a \, \vec{u}_{2y} - \sqrt{3} \, a \, \vec{u}_{2x} \\ B' = \vec{u}_{1y} - \sqrt{3} \, \vec{u}_{1x} \\ C' = y_0 - \sqrt{3} \, x_0 \\ B'^2 - 4 A' C' = 0$

Solving using iteration or search yields the parameters. I used a minimization routine which gave a pretty good result, although not quite as good as a multi-variate Newton-Raphson would give. The focal length is equal to $\large{\frac{1}{4 a}}$ , and it turns out to be about $0.58$ .

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Outputs
errsum = 0.00724383207176

x0 = 1.52864800995

y0 = 0.854106555955

theta (degrees) = 120.513963499 

a = 0.428939601063

alpha1 = 2.00889420741

alpha2 = -1.98812144104

f = 0.582832639795