This solution is much more efficient than the accepted answer. It’s execution time is logarithmic (while accepted answer has linear complexity).
var findYatXbyBisection = function(x, path, error){
var length_end = path.getTotalLength()
, length_start = 0
, point = path.getPointAtLength((length_end + length_start) / 2) // get the middle point
, bisection_iterations_max = 50
, bisection_iterations = 0
error = error || 0.01
while (x < point.x - error || x > point.x + error) {
// get the middle point
point = path.getPointAtLength((length_end + length_start) / 2)
if (x < point.x) {
length_end = (length_start + length_end)/2
} else {
length_start = (length_start + length_end)/2
}
// Increase iteration
if(bisection_iterations_max < ++ bisection_iterations)
break;
}
return point.y
}