You've probably seen this shape before. It is several straight lines whose outline seems to be curved. Connect the following pairs of points with segments:
(1,0) and (0,8)
(2,0) and (0,7)
(3,0) and (0,6)
(8,0) and (0,1)
The 8 segments seem to form the outline a curve, but what curve? Is it a circle or maybe one branch of a hyperbola? Some other conic? Does it's equation have some other nice form?
I used to do this with coloured threads on cardboard as a child. It was always a toss-up whether to complete the figure with the final two threads running straight up and across.
They turn out to be the key to the first part of the problem. In the given example, the last two threads will be from the origin to (9,0) and (0,9). Perpendiculars from the midpoints of these two threads will meet at (4.5,4.5). Indeed perpendiculars from the midpoints of all of the threads will meet at the same point. So the curve is by definition a parabola.
The general form of the equation is going to be a bit messy because the parabola is tilted by 45 degrees. To avoid unnecessary terms, let's start by rotating it so it opens upwards.
The math; let the focus of the parabola be= (a,b), and the directrix = k. Then the equation in terms of x and y will be: y=1/(2(b-k)) *(x-a)^2+1/2(b+k)
Now for the general case in the problem, with lines (1,0),(0,p) to (p,0)(0,1), we add one to p and divide by 2 to find the focus,(p+1)/2,(p+1)/2; doing the same as we did with the last two threads above.
To rotate the parabola we need to recompute k to make the opening point upwards. The directrix starts on the origin. The distance from the focus to the origin is (2*((p+1)/2)^2)^(1/2), or just (p+1)/sqrt(2), but because we rotated the figure for ease of handling, k becomes ((p+1)/2-(p+1)/sqrt(2)), always negative.
so substituting, y=1/(2((p+1)/2-((p+1)/2-(p+1)/sqrt(2)))) *(x-(p+1)/2)^2+1/2((p+1)/2+((p+1)/2-(p+1)/sqrt(2)))
e.g., for p=16,
y = x^2/(17 sqrt(2)) - x/sqrt(2) - 17/(4 sqrt(2)) + 17/2 ≈ 0.0415945 x^2 - 0.707107 x + 5.4948
Edited on April 19, 2017, 12:01 pm
Posted by broll
on 2017-04-19 11:49:05