平面椭圆参数方程推导

原创已于 2025-06-02 03:22:07 修改 · 871 阅读

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#线性代数 #matlab

于 2024-12-06 21:48:33 首次发布

平面椭圆参数方程推导

$ax^2+bxy+cy^2+dx+ey+f=0$

可以看作一个由一个 $Ax^2+By^2+C=0$ 的椭圆逆时针旋转 $θ\theta$ 再向右平移 $x_0$ ,向上平移 $y_0$ 得到

其中，不妨令 $0<A<B,θ∈[−π2,π2]0<A<B,\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]$ ，显然 $C < 0$

逆时针旋转 $θ\theta$ 后，方程变为
$A(x\cos(\theta)+y\sin(\theta))^2+B(-x\sin(\theta)+y\cos(\theta))^2+C =0$

%验证
A=1;B=4;C=-4;
thetas = deg2rad([-90,-60,-30,30,60,90]);
for ii = 1:1:length(thetas);
	subplot(1,length(thetas),ii)
	theta = thetas(ii);
	c = cos(theta);
    s = sin(theta);
    fa = [num2str(A),'*','(x*',num2str(c),'+y*',num2str(s),')^2'];
    fb = [num2str(B),'*','(-x*',num2str(s),'+y*',num2str(c),')^2'];
    f = [fa,'+',fb,'+',num2str(C)];
    ezplot(f);
    axis equal
end

展开后
$A(x\cos(\theta)+y\sin(\theta))^2+B(-x\sin(\theta)+y\cos(\theta))^2+C =0$
也即
$x^2(A\cos^2\theta+B\sin^2\theta)+xy((A-B)\sin(2\theta))+y^2(A\sin^2\theta+B\cos^2\theta)+C = 0$
向右平移 $x_0$ ,向上平移 $y_0$ 后

$(x-x_0)^2(A\cos^2\theta+B\sin^2\theta)+ (x-x_0)(y-y_0)((A-B)\sin(2\theta))\\+(y-y_0)^2(A\sin^2\theta+B\cos^2\theta)+C = 0$

$(x-x_0)^2(A\cos^2\theta+B\sin^2\theta) \\=x^2(A\cos^2\theta+B\sin^2\theta)\\-x(2x_0(A\cos^2\theta+B\sin^2\theta))\\+{x_0}^2(A\cos^2\theta+B\sin^2\theta)$

$(x-x_0)(y-y_0)((A-B)\sin(2\theta))\\ =xy((A-B)\sin(2\theta))\\ -x(y_0(A-B)\sin(2\theta))\\ -y(x_0(A-B)\sin(2\theta))\\ +x_0y_0((A-B)\sin(2\theta))$

$(y-y_0)^2(A\sin^2\theta+B\cos^2\theta) \\=y^2(A\sin^2\theta+B\cos^2\theta)\\-y(2y_0(A\sin^2\theta+B\cos^2\theta))\\+{y_0}^2(A\sin^2\theta+B\cos^2\theta)$

已知量 $a, b, c, d, e, f$ 与未知量 $A,B,θ,x0,y0,CA,B,\theta,x_0,y_0,C$ 的关系
$A\cos^2\theta+B\sin^2\theta\\ b = (A-B)\sin(2\theta)\\ c = A\sin^2\theta+B\cos^2\theta\\ d = -2ax_0-by_0\\ e = -bx_0-2cy_0\\ f = a{x_0}^2+b{x_0}{y_0}+c{y_0}^2+C$
求解 $A,B,θA,B,\theta$
$A+B\\ a-c = (A-B)\cos(2\theta)\\ b = (A-B)\sin(2\theta)\\$
$A<B,θ∈[−π2,π2]A<B,\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]$ ，所以
$\sqrt{(a-c)^2+b^2} = B-A\\ A = \frac{a+c - \sqrt{(a-c)^2+b^2}}{2}\\ B = \frac{a+c + \sqrt{(a-c)^2+b^2}}{2}\\$

if(a-c<0)
    theta = 1/2*asin(b/(A-B));
elseif(b<0)
    theta = pi/2 - 1/2*asin(b/(A-B));
else
    theta = -pi/2 - 1/2*asin(b/(A-B));
end

求解x0,y0
$x_0 = (2cd-be)/(b^2 - 4ac)\\ y_0 = (2ae-bd)/(b^2 - 4ac)$
求解 $C$
$C = f - a{x_0}^2-b{x_0}{y_0}-c{y_0}^2$

$(x-x_0)^2(A\cos^2\theta+B\sin^2\theta)+ (x-x_0)(y-y_0)((A-B)\sin(2\theta))\\+(y-y_0)^2(A\sin^2\theta+B\cos^2\theta)+C = 0$

令 $x_1 = x-x_0,y_1=y-y_0$ ，也即 $x = x_1+x_0,y= y_0 + y_1$
$x_1^2(A\cos^2\theta+B\sin^2\theta)+x_1y_1((A-B)\sin(2\theta))+y_1^2(A\sin^2\theta+B\cos^2\theta)+C = 0$

$\begin{bmatrix} x_1&y_1 \end{bmatrix} \begin{bmatrix} A\cos^2\theta+B\sin^2\theta&(A-B)\sin(\theta)\cos(\theta)\\ (A-B)\sin(\theta)\cos(\theta)&A\sin^2\theta+B\cos^2\theta\\ \end{bmatrix} \begin{bmatrix} x_1\\y_1 \end{bmatrix}+C = 0$

$(\begin{bmatrix} \cos(\theta)&\sin(\theta)\\ -\sin(\theta)&\cos(\theta)\\ \end{bmatrix} \begin{bmatrix} x_1\\y_1 \end{bmatrix})^\mathrm{T} \begin{bmatrix} A&0\\0&B \end{bmatrix}\begin{bmatrix} \cos(\theta)&\sin(\theta)\\ -\sin(\theta)&\cos(\theta)\\ \end{bmatrix}\begin{bmatrix} x_1\\y_1 \end{bmatrix}+C = 0$

令 $x2=x1cos⁡(θ)+y1sin⁡(θ),y2=−x1sin⁡(θ)+y1cos⁡(θ)x_2 = x_1\cos(\theta)+y_1\sin(\theta),y_2 = -x_1\sin(\theta)+y_1\cos(\theta)$ ，也即 $x1=x2cos⁡(θ)−y2sin⁡(θ),y1=x2sin⁡(θ)+y2cos⁡(θ)x_1 = x_2\cos(\theta)-y_2\sin(\theta),y_1 = x_2\sin(\theta)+y_2\cos(\theta)$
$Ax_2^2 + By_2^2+C = 0$
参数方程
$x_2 = \sqrt{\frac{-C}{A}}\cos(t)\\ y_2 = \sqrt{\frac{-C}{B}}\sin(t)$

$x_1 = \sqrt{\frac{-C}{A}}\cos(\theta)\cos(t)-\sqrt{\frac{-C}{B}}\sin(\theta)\sin(t)\\ y_1 = \sqrt{\frac{-C}{A}}\sin(\theta)\cos(t)+\sqrt{\frac{-C}{B}}\cos(\theta)\sin(t)$

最终参数方程
$\sqrt{\frac{-C}{A}}\cos(\theta)\cos(t)-\sqrt{\frac{-C}{B}}\sin(\theta)\sin(t)+x_0\\ y= \sqrt{\frac{-C}{A}}\sin(\theta)\cos(t)+\sqrt{\frac{-C}{B}}\cos(\theta)\sin(t)+y_0$

已知量	未知量	表达式
a	A	$a+c−(a−c)2+b22\frac{a+c - \sqrt{(a-c)^2+b^2}}{2}$
b	B	$a+c+(a−c)2+b22\frac{a+c + \sqrt{(a-c)^2+b^2}}{2}$
c	C	$f - a{x_0}^2-b{x_0}{y_0}-c{y_0}^2$
d	$x_0$	$2cd-be)/(b^2 - 4ac)$
e	$y_0$	$2ae-bd)/(b^2 - 4ac)$
f	$θ\theta$	$见上述 ma tl ab 代码$

验证代码

clf;
a = rand(1);
b = rand(1);
c = rand(1);
d = 0.5*rand(1)-0.25;
e = 0.5*rand(1)-0.25;
f = 100*rand(1)-100;
func = @(x,y)(a*x*x+b*x*y+c*y*y+d*x+e*y+f);
fimplicit(func);
axis([-50,50,-50,50]);
pause(1);
aa = (a+c - sqrt((a-c)^2+b^2))/2;
bb = (a+c + sqrt((a-c)^2+b^2))/2;
x0 = (2*c*d-b*e)/(b*b-4*a*c);
y0 = (2*a*e-b*d)/(b*b-4*a*c);
cc = f-a*x0*x0-b*x0*y0-c*y0*y0;
if(a-c<0)
    theta = 1/2*asin(b/(aa-bb));
elseif(b<0)
    theta = pi/2 - 1/2*asin(b/(aa-bb));
else
    theta = -pi/2 - 1/2*asin(b/(aa-bb));
end

xt = @(t)(sqrt(-cc/aa)*cos(theta)*cos(t)-sqrt(-cc/bb)*sin(theta)*sin(t)+x0);
yt = @(t)(sqrt(-cc/aa)*sin(theta)*cos(t)+sqrt(-cc/bb)*cos(theta)*sin(t)+y0);
tt = linspace(0,2*pi,100);
xplot = zeros(1,100);
yplot = zeros(1,100);
for ii = 1:1:100
    xplot(ii) = xt(tt(ii));
    yplot(ii) = yt(tt(ii));
end
hold on;plot(xplot,yplot);