我有一个关于抛物线曲线与指定点相交的方程，这里指用户在图表上单击的位置。
// this would typically be mouse coords on the graph
var _target:Point = new Point(100, 50);
public static function plot(x:Number, target:Point):Number{
return (x * x) / target.x * (target.y / target.x);
}
这将生成如下图所示的图形：
startX:Number, startY:Number, endX:Number, endY:Number
我需要找出这条曲线与这些线段（A）相交的位置：
如果有帮助的话，startX总是小于endX
我感觉有一个相当简单的方法来做到这一点，但我不知道要搜索什么，也不太精通“适当”的数学，所以实际的代码示例将非常受欢迎。
更新：
我已经解决了交点的问题，但我的解决方案给出了错误侧面的坐标。
用A和B替换我的目标坐标，得到了该图的方程：
(x * x) / A * (B/A)
// this simplifies down to:
(B * x * x) / (A * A)
// which i am the equating to the line's equation
(B * x * x) / (A * A) = m * x + b
// i run this through wolfram alpha (because i have no idea what i'm doing) and get:
(A * A * m - A * Math.sqrt(A * A * m * m + 4 * b * B)) / (2 * B)
这是一个正确的答案，但我想要第二种可能的变化。我已经通过在计算前将m乘以-1来进行修正，并在最后一次计算返回的x值上执行相同的操作来纠正这个问题，但感觉像是一个hack。
解决方案：
public static function intersectsSegment(targetX:Number, targetY:Number, startX:Number, startY:Number, endX:Number, endY:Number):Point {
// slope of the line
var m:Number = (endY - startY) / (endX - startX);
// where the line intersects the y-axis
var b:Number = startY - startX * m;
// solve the two variatons of the equation, we may need both
var ix1:Number = solve(targetX, targetY, m, b);
var ix2:Number = solveInverse(targetX, targetY, m, b);
var intersection1:Point;
var intersection2:Point;
// if the intersection is outside the line segment startX/endX it's discarded
if (ix1 > startX && ix1 < endX) intersection1 = new Point(ix1, plot(ix1, targetX, targetY));
if (ix2 > startX && ix2 < endX) intersection2 = new Point(ix2, plot(ix2, targetX, targetY));
// somewhat fiddly code to return the smallest set intersection
if (intersection1 && intersection2) {
// return the intersection with the smaller x value
return intersection1.x < intersection2.x ? intersection1 : intersection2;
} else if (intersection1) {
return intersection1;
}
// this effectively means that we return intersection2 or if that's unset, null
return intersection2;
}
private static function solve(A:Number, B:Number, m:Number, b:Number):Number {
return (m + Math.sqrt(4 * (B / (A * A)) * b + m * m)) / (2 * (B / (A * A)));
}
private static function solveInverse(A:Number, B:Number, m:Number, b:Number):Number {
return (m - Math.sqrt(4 * (B / (A * A)) * b + m * m)) / (2 * (B / (A * A)));
}
public static function plot(x:Number, targetX:Number, targetY:Number):Number{
return (targetY * x * x) / (targetX * targetX);
}