Simple Question Complicated Answer - 2

Label the three points $A(0, 0)$ , $B(3, 3\sqrt{3})$ , and $C(4, 0)$ , and draw $BC$ . By the distance formula, the sides of $\triangle ABC$ are $AC = 4$ , $AB = 6$ , and $BC = 2 \sqrt{7}$ . Now rotate and shift the diagram so that the parabola has a vertical axis and a vertex at $(0, 0)$ , for an equation of $y = ax^2$ , while still preserving the focal distance and triangle lengths of $\triangle A'B'C'$ .

Let $B'$ be $(u, au^2)$ and $C'$ be $(v, av^2)$ . The slope at $B'$ is $y' = 2au$ so its tangent line is $y = 2aux - au^2$ . Likewise, the tangent line at $C'$ is is $y = 2avx - av^2$ . These tangent lines intersect at $A'(\frac{1}{2}u + \frac{1}{2}v, auv)$ .

Using the distance formula for each side of $\triangle A'B'C'$ gives

$B'C'^2 = (u - v)^2 + (au^2 - av^2)^2 = 28$

$A'C'^2 = (\frac{1}{2}u - \frac{1}{2}v)^2 + (auv - av^2)^2 = 16$

$A'B'^2 = (\frac{1}{2}u - \frac{1}{2}v)^2 + (auv - au^2)^2 = 36$

which for $a > 0$ has a solution of $a = \frac{19\sqrt{19}}{216}$ .

Therefore, the focal distance is $p = \frac{1}{4 \cdot \frac{19\sqrt{19}}{216}} = \frac{54 \sqrt{19}}{361} \approx \boxed{0.652}$ .

This problem lends itself to the Bezier curve method, which states the the parametric equation of the given parabola is,

$p(t) = p_0 (1 - t)^2 + 2 p_1 t (1 - t) + p_2 t^2$

where $p_0$ and $p_2$ are the tangent points of the parabola and $p_1$ is the intersection point of the two tangent lines.

So we can take $p_0 = (4, 0)$ , and $p_2 = ( 3, 3 \sqrt{3} )$ and $p_1 = (0, 0)$ . Substituting this,

$p(t) = ( 4 (1 - t)^2 + 3 t^2 , 3 \sqrt{3} t^2 )$

$= ( 4 - 8 t + 7 t^2 , 3 \sqrt{3} t^2 )$

$= q_0 + q_1 t + q_2 t^2$

where $q_0 = (4, 0) , q_1 = (-8, 0 ), q_2 = (7, 3 \sqrt{3} )$

Now we'll apply a small trick to orthogonalize the two axes spanning $p(t)$ . Since the parameter $t$ can be any real number, we can, just as well, use $(t - t_0)$ instead of $t$ , where $t_0$ is some fixed real number. Thus

$p(t) = q_0 + q_1 (t - t_0) + q_2 (t - t_0)^2$

$= (q_0 - q_1 t_0 + q_2 t_0^2 ) + ( q_1 - 2 q_2 t_0 ) t + q_2 t^2$

Now, we'll select $t_0$ such that,

$(q_1 - 2 q_2 t_0) \cdot q_2 = 0$

which is always possible by choosing $t_0 = \dfrac{1}{2} \dfrac{ (q_1 . q_2) }{ (q_2 . q_2) }$ , then

$p(t) = w_0 + w_1 t + w_2 t^2$

where,

$w_0 = q_0 - q_1 t_0 + q_2 t_0^2$

$w_1 = q_1 - 2 q_2 t_0$

$w_2 = q_2$

and with the above value of $t_0$ , we have $w_1$ orthogonal to $w_2$ . If we now define unit vectors along $w_1$ and $w_2$ as follows

$u_1 = w_1 / | w_1 |$ and $u_2 = w_2 / | w_2 |$ , then

$p(t) = w_0 + | w_1 | u_1 t + | w_2 | u_2 t^2$

since $u_1$ and $u_2$ are unit vectors and are orthogonal, they specify a reference frame, hence

$p(t) - w_0 = [u_1 , u_2 ](x, y)$

with $x = | w_1 | t$ , and $y = | w_2 | t^2$

thus $(\dfrac{x}{ | w_1 |} )^2 = \dfrac{y}{ | w_2 | }$

is our equation of the parabola; it follows that $y ( \dfrac{| w_1 |^2 }{ | w_2 | } ) = x^2$

Comparing this with the standard equation of a parabola, which is $4 p y = x^2$ , it follows that,

$4 p = \dfrac{ | w_1 |^2 }{ | w_2 | }$

which imples that $p = \dfrac{1}{4} \dfrac{ | w_1 |^2 }{ | w_2 | }$

The rest is straight forward computations.