For all t in the interval 0≤t≤1, a thread is pinned from the point (0, t) on the y-axis to the corresponding point (1-t, 0) on the x-axis.
Under what curve do the threads lie?
The lines are all parametrized in t and are of the form y = x*t/(t-1) + t. These lines are all tangent to the curve.
Now solve the line equation for t:
y = x*t/(t-1) + t
0 = t*(t-1) + x*t - y*(t-1)
0 = t^2 - t + x*t - y*t + y
0 = t^2 + (x-y-1)*t + (y)
Then quadratic formula gets us
t = ((-x+y+1) +/- sqrt[(x-y-1)^2-4y]) / 2
The important part here is the discriminant. When the line is tangent to our curve we will have this equal to 0. Thus the curve is (x-y-1)^2 - 4y = 0.
Expand into standard form to get:
x*(x-y-1) - y*(x-y-1) - (x-y-1) - 4y = 0
x^2 - xy - x - xy + y^2 + y - x + y + 1 - 4y = 0
x^2 - 2xy + y^2 - 2x - 2y + 1 = 0.
What kind of curve is it? Find the discriminant of the curve D=b^2-4ac, which for this curve is (-2)^2-4*1*1=0. A value of zero means this curve is a parabola.
Edited on January 3, 2025, 5:32 pm
Solution
