3 ms·
Sorry, I accidentally down-voted you when I tried to up-vote you. This answer is absolutely correct. Here is my very similar derivation: All lines go through (
by jules 14y ago
Sorry, I accidentally down-voted you when I tried to up-vote you. This answer is absolutely correct. Here is my very similar derivation:
All lines go through (t,0) and (0,1-t). Now take two such lines, one with constant t and another with constant s and find their intersection. We have:
y = 1-t - (1-t)/t * x
y = 1-s - (1-s)/s * x
So:
1-t - (1-t)/t * x = 1-s - (1-s)/s
s-t = (s-t)/(st) * x
So:
x = st
The points on the curve will be generated by the intersection of two almost adjacent lines (this is easy to see geometrically), so we take s =~ t. Then we have:
x = t^2
y = (1-t)^2
So we have:
y = (1 - sqrt(x))^2
Or, my favorite form:
sqrt(x) + sqrt(y) = 1