1 = 4a
Z
x=t
1 + cosh 2θdθ = [ x=0
sinh2θ x=t θ + ] 4a 8a x=0
From (1), we can rewrite this as √ arsinh(2at) + 2at 1 + 4a2 t2 s(t) = 4a √ And since it can be shown that arsinh(v) = ln(v + 1 + v 2 ), we have √ √ ln(2at + 1 + 4a2 t2 ) t 1 + 4a2 t2 s(t) = + 4a 2
(2)
This is something we will return to later. Now if the parabola is standing up, visualizing the situation again, it is clear that if we define an angle φ to be the angle between the tangent line to the parabola at P = (t, at2 ), and the secant line from (0, 0) to P on the parabola, then if the parabola is later tilted such that the x axis is tangent to P , then the vertex will lie at the point x(t) = s(t) − R cos (φ(t))
(3)
y(t) = R sin (φ(t))
(4)
Where R is the distance √ between P and √ the vertex. 2 2 4 It is clear that R = t + a t = |t| 1 + a2 t2 , and s(t) is already known. So, now all that is needed is an expression φ(t). If we construct two vectors, one in the direction of the tangent line mentioned above, and one in the direction of the secant line, and dot the two, it is possible to find the angle between them by the theorem a · b = |ab| cos φ. It is quite simple to derive the tangent and secant√ vectors to be < 1, 2at > √ and < t, at2 > with magnitudes 1 + 4a2 t2 and t 1 + a2 t2 , respectively. The dot product of the two vectors is also trivial to find, and can be expressed as t + 2a2 t3 . Now, using the dot product theorem, we can solve for φ. p p t 1 + a2 t2 1 + 4a2 t2 cos φ = t + 2a2 t3 1 + 2a2 t2 cos φ = p (1 + a2 t2 )(1 + 4a2 t2 ) φ = arccos p
1 + 2a2 t2 (1 + a2 t2 )(1 + 4a2 t2 ) 2
(5)