標準形式以外的選讀拓展
方程 x y = 1 xy=1 x y = 1 x 2 + y 2 = 0 x^2+y^2=0 x 2 + y 2 = 0
焦點準線條件解釋了二次方程的來源。設焦點為 ( p , q ) (p,q) ( p , q ) u x + v y + w = 0 ux+vy+w=0 ux + v y + w = 0 e > 0 e\gt0 e > 0 u 2 + v 2 > 0 u^2+v^2\gt0 u 2 + v 2 > 0
( x − p ) 2 + ( y − q ) 2 = e ∣ u x + v y + w ∣ u 2 + v 2 . \sqrt{(x-p)^2+(y-q)^2}
=e\frac{|ux+vy+w|}{\sqrt{u^2+v^2}}. ( x − p ) 2 + ( y − q ) 2 = e u 2 + v 2 ∣ ux + v y + w ∣ . 兩邊非負，故平方可逆，再乘以 u 2 + v 2 u^2+v^2 u 2 + v 2
對實系數寫
f ( x , y ) = A x 2 + B x y + C y 2 + D x + E y + F . f(x,y)=Ax^2+Bxy+Cy^2+Dx+Ey+F. f ( x , y ) = A x 2 + B x y + C y 2 + D x + E y + F . 實軌跡是所有滿足 f ( x , y ) = 0 f(x,y)=0 f ( x , y ) = 0 A x 2 + B x y + C y 2 Ax^2+Bxy+Cy^2 A x 2 + B x y + C y 2 Δ = B 2 − 4 A C \Delta=B^2-4AC Δ = B 2 − 4 A C A = B = C = 0 A=B=C=0 A = B = C = 0
全式乘以非零常數不改變軌跡。座標旋轉與平移也保持距離及角度；它們改變同一集合的描述，不改變幾何類型。先旋轉消去混合項，再平移，便能看清哪些常數項與一次項仍決定形狀。
旋轉座標軸
將座標軸逆時針旋轉 θ \theta θ c = cos θ c=\cos\theta c = cos θ s = sin θ s=\sin\theta s = sin θ ( X , Y ) (X,Y) ( X , Y ) ( c , s ) (c,s) ( c , s ) ( − s , c ) (-s,c) ( − s , c )
x = c X − s Y , y = s X + c Y ; X = c x + s y , Y = − s x + c y . x=cX-sY,\qquad y=sX+cY;
\qquad X=cx+sy,\qquad Y=-sx+cy. x = c X − s Y , y = s X + c Y ; X = c x + sy , Y = − s x + cy . 利用 c 2 + s 2 = 1 c^2+s^2=1 c 2 + s 2 = 1
代入後新方程的系數為
A ′ = A c 2 + B c s + C s 2 , C ′ = A s 2 − B c s + C c 2 , B ′ = B cos 2 θ − ( A − C ) sin 2 θ , D ′ = D c + E s , E ′ = − D s + E c , F ′ = F . \begin{aligned}
A'&=Ac^2+Bcs+Cs^2, & C'&=As^2-Bcs+Cc^2,\\
B'&=B\cos2\theta-(A-C)\sin2\theta,\\
D'&=Dc+Es, & E'&=-Ds+Ec,\qquad F'=F.
\end{aligned} A ′ B ′ D ′ = A c 2 + B cs + C s 2 , = B cos 2 θ − ( A − C ) sin 2 θ , = D c + E s , C ′ E ′ = A s 2 − B cs + C c 2 , = − D s + E c , F ′ = F . 總能選擇實角使 B ′ = 0 B'=0 B ′ = 0
A ′ + C ′ = A + C , B ′ 2 − 4 A ′ C ′ = B 2 − 4 A C . A'+C'=A+C,\qquad B'^2-4A'C'=B^2-4AC. A ′ + C ′ = A + C , B ′2 − 4 A ′ C ′ = B 2 − 4 A C . 證明: 先找出要消去的系數，再選擇角度
目標是消去 X Y XY X Y A x 2 Ax^2 A x 2 B x y Bxy B x y C y 2 Cy^2 C y 2 − 2 A c s + B ( c 2 − s 2 ) + 2 C c s -2Acs+B(c^2-s^2)+2Ccs − 2 A cs + B ( c 2 − s 2 ) + 2 C cs B ′ B' B ′ B = 0 B=0 B = 0 θ = 0 \theta=0 θ = 0 B ≠ 0 B\ne0 B = 0 A = C A=C A = C θ = π / 4 \theta=\pi/4 θ = π /4 tan 2 θ = B / ( A − C ) \tan2\theta=B/(A-C) tan 2 θ = B / ( A − C ) 2 θ 2\theta 2 θ − π / 2 -\pi/2 − π /2 π / 2 \pi/2 π /2 B ′ = 0 B'=0 B ′ = 0 A = C A=C A = C
將 A ′ A' A ′ C ′ C' C ′ B c s Bcs B cs − B c s -Bcs − B cs A ′ − C ′ = ( A − C ) cos 2 θ + B sin 2 θ A'-C'=(A-C)\cos2\theta+B\sin2\theta A ′ − C ′ = ( A − C ) cos 2 θ + B sin 2 θ B ′ B' B ′
B ′ 2 + ( A ′ − C ′ ) 2 = B 2 + ( A − C ) 2 . B'^2+(A'-C')^2=B^2+(A-C)^2. B ′2 + ( A ′ − C ′ ) 2 = B 2 + ( A − C ) 2 . 再減去已經證明不變的平方 ( A ′ + C ′ ) 2 = ( A + C ) 2 (A'+C')^2=(A+C)^2 ( A ′ + C ′ ) 2 = ( A + C ) 2
之後的平移不改變二次系數，因為展開平移後的平方只新增一次項和常數項。因此 Δ \Delta Δ r r r r r r r 2 r^2 r 2
互動 · 探索
旋轉座標軸，保持橢圓不動
旋轉座標軸，找出 XY 係數為零的角度。
新座標 5 X 2 + 6 X Y + 5 Y 2 = 8 5X^2 + 6XY + 5Y^2 = 8 5 X 2 + 6 X Y + 5 Y 2 = 8 XY 係數 ≈ 6 不成立：交叉項係數為零
B = 6 cos ( 2 θ ) B=6\cos(2\theta) B = 6 cos ( 2 θ ) B = 0 B=0 B = 0 θ = 45 ∘ \theta=45^\circ θ = 4 5 ∘
要留意 固定橢圓為 5 x 2 + 6 x y + 5 y 2 = 8 5x^2+6xy+5y^2=8 5 x 2 + 6 x y + 5 y 2 = 8 X , Y X,Y X , Y x = X cos θ − Y sin θ x=X\cos\theta-Y\sin\theta x = X cos θ − Y sin θ y = X sin θ + Y cos θ y=X\sin\theta+Y\cos\theta y = X sin θ + Y cos θ 45 ∘ 45^\circ 4 5 ∘ 8 X 2 + 2 Y 2 = 8 8X^2+2Y^2=8 8 X 2 + 2 Y 2 = 8
配方並保留全部情形
旋轉之後重新命名系數，寫成
α X 2 + γ Y 2 + d X + e Y + f = 0. \alpha X^2+\gamma Y^2+dX+eY+f=0. α X 2 + γ Y 2 + d X + e Y + f = 0. 此處的小寫 e e e α , γ \alpha,\gamma α , γ
若二次部分的秩為二，令 h = − d / ( 2 α ) h=-d/(2\alpha) h = − d / ( 2 α ) k = − e / ( 2 γ ) k=-e/(2\gamma) k = − e / ( 2 γ ) m = d 2 / ( 4 α ) + e 2 / ( 4 γ ) − f m=d^2/(4\alpha)+e^2/(4\gamma)-f m = d 2 / ( 4 α ) + e 2 / ( 4 γ ) − f U = X − h U=X-h U = X − h V = Y − k V=Y-k V = Y − k
α U 2 + γ V 2 = m . \alpha U^2+\gamma V^2=m. α U 2 + γ V 2 = m . 若 α , γ \alpha,\gamma α , γ − 1 -1 − 1 m m m m > 0 m\gt0 m > 0 m = 0 m=0 m = 0 m < 0 m\lt0 m < 0 m ≠ 0 m\ne0 m = 0 m = 0 m=0 m = 0
若秩為一，必要時交換座標角色，使 α ≠ 0 \alpha\ne0 α = 0 γ = 0 \gamma=0 γ = 0
α U 2 + e Y + g = 0 , U = X + d 2 α , g = f − d 2 4 α . \alpha U^2+eY+g=0,\qquad
U=X+\frac{d}{2\alpha},\qquad g=f-\frac{d^2}{4\alpha}. α U 2 + e Y + g = 0 , U = X + 2 α d , g = f − 4 α d 2 . 若 e ≠ 0 e\ne0 e = 0 e = 0 e=0 e = 0 r = − g / α r=-g/\alpha r = − g / α r > 0 r\gt0 r > 0 r = 0 r=0 r = 0 r < 0 r\lt0 r < 0
若秩為零，方程為 d X + e Y + f = 0 dX+eY+f=0 d X + e Y + f = 0 ( d , e ) ≠ ( 0 , 0 ) (d,e)\ne(0,0) ( d , e ) = ( 0 , 0 ) d = e = f = 0 d=e=f=0 d = e = f = 0
同號情形的依據是正系數平方和總非負，且僅在兩個變量都為零時為零。
m m m m m m m / α \sqrt{m/\alpha} m / α m / γ \sqrt{m/\gamma} m / γ α U 2 − β V 2 = m \alpha U^2-\beta V^2=m α U 2 − β V 2 = m α , β > 0 \alpha,\beta\gt0 α , β > 0 ( α U − β V ) ( α U + β V ) = 0 (\sqrt\alpha U-\sqrt\beta V)(\sqrt\alpha U+\sqrt\beta V)=0 ( α U − β V ) ( α U + β V ) = 0
秩一且 e ≠ 0 e\ne0 e = 0 V = Y + g / e V=Y+g/e V = Y + g / e α U 2 = − e V \alpha U^2=-eV α U 2 = − e V − e / α -e/\alpha − e / α Y 2 Y^2 Y 2 e Y eY e Y Y Y Y e = 0 e=0 e = 0 U U U Y Y Y
反例模式
判別式為正並不能排除相交直線 方程 x 2 − y 2 = 0 x^2-y^2=0 x 2 − y 2 = 0 Δ = 4 > 0 \Delta=4\gt0 Δ = 4 > 0 ( x − y ) ( x + y ) = 0 (x-y)(x+y)=0 ( x − y ) ( x + y ) = 0 y = x y=x y = x y = − x y=-x y = − x x 2 − y 2 = 1 x^2-y^2=1 x 2 − y 2 = 1
正判別式所確定的是二次部分不定號，配方後仍須檢查剩餘常數是否為零。同樣，負判別式容許橢圓、單點或空集；零判別式則須繼續檢查保留的一次項與秩。
圓要求 A = C ≠ 0 A=C\ne0 A = C = 0 B = 0 B=0 B = 0 B 2 + ( A − C ) 2 = 0 B^2+(A-C)^2=0 B 2 + ( A − C ) 2 = 0 A A A ( − D / ( 2 A ) , − E / ( 2 A ) ) (-D/(2A),-E/(2A)) ( − D / ( 2 A ) , − E / ( 2 A ))
R 2 = D 2 + E 2 4 A 2 − F A . R^2=\frac{D^2+E^2}{4A^2}-\frac FA. R 2 = 4 A 2 D 2 + E 2 − A F . 只有 R 2 > 0 R^2\gt0 R 2 > 0
分類例題
分類下列方程，並確定其軸：
5 x 2 − 4 x y + 8 y 2 + 4 5 x − 16 5 y + 4 = 0. 5x^2-4xy+8y^2+4\sqrt5x-16\sqrt5y+4=0. 5 x 2 − 4 x y + 8 y 2 + 4 5 x − 16 5 y + 4 = 0. 此處 Δ = − 144 \Delta=-144 Δ = − 144 c = 2 / 5 c=2/\sqrt5 c = 2/ 5 s = 1 / 5 s=1/\sqrt5 s = 1/ 5 tan 2 θ = 4 / 3 \tan2\theta=4/3 tan 2 θ = 4/3 x = ( 2 X − Y ) / 5 x=(2X-Y)/\sqrt5 x = ( 2 X − Y ) / 5 y = ( X + 2 Y ) / 5 y=(X+2Y)/\sqrt5 y = ( X + 2 Y ) / 5
4 X 2 + 9 Y 2 − 8 X − 36 Y + 4 = 0 ⟺ 4 ( X − 1 ) 2 + 9 ( Y − 2 ) 2 = 36. 4X^2+9Y^2-8X-36Y+4=0
\quad\Longleftrightarrow\quad
4(X-1)^2+9(Y-2)^2=36. 4 X 2 + 9 Y 2 − 8 X − 36 Y + 4 = 0 ⟺ 4 ( X − 1 ) 2 + 9 ( Y − 2 ) 2 = 36. 右側為正，現在才確認是實橢圓。旋轉座標中的中心為 ( 1 , 2 ) (1,2) ( 1 , 2 ) ( 0 , 5 ) (0,\sqrt5) ( 0 , 5 ) ( 2 , 1 ) / 5 (2,1)/\sqrt5 ( 2 , 1 ) / 5 3 3 3 ( − 1 , 2 ) / 5 (-1,2)/\sqrt5 ( − 1 , 2 ) / 5 2 2 2
例題 11.4.5
（例 2：一次項區分拋物線與平行直線）
考慮 x 2 + 2 x y + y 2 − 2 x + 2 y = 0 x^2+2xy+y^2-2x+2y=0 x 2 + 2 x y + y 2 − 2 x + 2 y = 0 Δ = 0 \Delta=0 Δ = 0 θ = π / 4 \theta=\pi/4 θ = π /4 x + y = 2 X x+y=\sqrt2X x + y = 2 X − 2 x + 2 y = 2 2 Y -2x+2y=2\sqrt2Y − 2 x + 2 y = 2 2 Y 2 X 2 + 2 2 Y = 0 2X^2+2\sqrt2Y=0 2 X 2 + 2 2 Y = 0 Y Y Y
對比 x 2 + 2 x y + y 2 − 2 = 0 x^2+2xy+y^2-2=0 x 2 + 2 x y + y 2 − 2 = 0 X 2 = 1 X^2=1 X 2 = 1 x + y = ± 2 x+y=\pm\sqrt2 x + y = ± 2 − 2 -2 − 2 0 0 0 x + y = 0 x+y=0 x + y = 0 2 2 2
例題 11.4.6
（例 3：相同定號二次部分的三種結果）
方程 x 2 + y 2 − 2 x + 4 y + c = 0 x^2+y^2-2x+4y+c=0 x 2 + y 2 − 2 x + 4 y + c = 0 Δ = − 4 \Delta=-4 Δ = − 4 ( x − 1 ) 2 + ( y + 2 ) 2 = 5 − c (x-1)^2+(y+2)^2=5-c ( x − 1 ) 2 + ( y + 2 ) 2 = 5 − c c = 1 c=1 c = 1 ( 1 , − 2 ) (1,-2) ( 1 , − 2 ) 2 2 2 c = 5 c=5 c = 5 ( 1 , − 2 ) (1,-2) ( 1 , − 2 ) c = 6 c=6 c = 6
二次部分的矩陣觀點
概念視角 代數
對稱矩陣記錄同一個二次表達式 定義欄向量 z = ( x , y ) T z=(x,y)^T z = ( x , y ) T Z = ( X , Y ) T Z=(X,Y)^T Z = ( X , Y ) T
Q = ( A B / 2 B / 2 C ) , O = ( c − s s c ) . Q=\begin{pmatrix}A&B/2\\B/2&C\end{pmatrix},\qquad
O=\begin{pmatrix}c&-s\\s&c\end{pmatrix}. Q = ( A B /2 B /2 C ) , O = ( c s − s c ) . 則 z T Q z = A x 2 + B x y + C y 2 z^TQz=Ax^2+Bxy+Cy^2 z T Q z = A x 2 + B x y + C y 2 z = O Z z=OZ z = O Z O T O = O O T = I O^TO=OO^T=I O T O = O O T = I
z T Q z = Z T ( O T Q O ) Z , Q ′ = O T Q O , Q = O Q ′ O T . z^TQz=Z^T(O^TQO)Z,\qquad Q'=O^TQO,
\qquad Q=OQ'O^T. z T Q z = Z T ( O T QO ) Z , Q ′ = O T QO , Q = O Q ′ O T . 最後一個等式由左乘 O O O O T O^T O T B ′ = 0 B'=0 B ′ = 0 Q ′ Q' Q ′
跡是對角元素之和 A + C A+C A + C A C − B 2 / 4 = − Δ / 4 AC-B^2/4=-\Delta/4 A C − B 2 /4 = − Δ/4 det O = 1 \det O=1 det O = 1 Q ′ = O T Q O Q'=O^TQO Q ′ = O T QO ( D , E ) O Z (D,E)OZ ( D , E ) O Z
切線要求接觸點為正則點
梯度 ∇ f = ( f x , f y ) \nabla f=(f_x,f_y) ∇ f = ( f x , f y ) 雙曲線的推導
已說明：沿曲線對方程求導，所得梯度與切線垂直。一般二次軌跡卻可能有交叉分支，甚至只有一個孤立點，故這裏必須保留梯度非零的條件。
定理 11.4.7
（二次軌跡在正則點處的切線公式）
設 P = ( x 0 , y 0 ) P=(x_0,y_0) P = ( x 0 , y 0 ) f ( P ) = 0 f(P)=0 f ( P ) = 0
∇ f ( P ) = ( 2 A x 0 + B y 0 + D , B x 0 + 2 C y 0 + E ) ≠ ( 0 , 0 ) . \nabla f(P)=(2Ax_0+By_0+D,\ Bx_0+2Cy_0+E)\ne(0,0). ∇ f ( P ) = ( 2 A x 0 + B y 0 + D , B x 0 + 2 C y 0 + E ) = ( 0 , 0 ) . 則切線為 ∇ f ( P ) ⋅ ( x − x 0 , y − y 0 ) = 0 \nabla f(P)\cdot(x-x_0,y-y_0)=0 ∇ f ( P ) ⋅ ( x − x 0 , y − y 0 ) = 0
A x 0 x + B 2 ( x 0 y + y 0 x ) + C y 0 y + D 2 ( x + x 0 ) + E 2 ( y + y 0 ) + F = 0. Ax_0x+\frac B2(x_0y+y_0x)+Cy_0y
+\frac D2(x+x_0)+\frac E2(y+y_0)+F=0. A x 0 x + 2 B ( x 0 y + y 0 x ) + C y 0 y + 2 D ( x + x 0 ) + 2 E ( y + y 0 ) + F = 0. 上述標準形式在梯度非零的點均有正則局部參數表示。非退化圓錐曲線已有前幾節的參數式，直線亦有常數非零切向量。若軌跡由兩條直線組成，梯度條件排除了交點，局部只會見到一條直線；重直線及孤立點上的梯度均為零。旋轉和平移則保留參數表示的正則性。
沿這樣的參數曲線對 f ( x ( t ) , y ( t ) ) = 0 f(x(t),y(t))=0 f ( x ( t ) , y ( t )) = 0 ∇ f ( P ) ⋅ ( x ′ ( t ) , y ′ ( t ) ) = 0 \nabla f(P)\cdot(x'(t),y'(t))=0 ∇ f ( P ) ⋅ ( x ′ ( t ) , y ′ ( t )) = 0
展開點法式，利用 f ( P ) = 0 f(P)=0 f ( P ) = 0 0 = 0 0=0 0 = 0 x 2 − y 2 = 0 x^2-y^2=0 x 2 − y 2 = 0
快速檢查
若 A = C A=C A = C A − C A-C A − C
解答 · 檢查 1 的答案 若 B = 0 B=0 B = 0 θ = 0 \theta=0 θ = 0 θ = π / 4 \theta=\pi/4 θ = π /4 B ′ = B cos ( π / 2 ) = 0 B'=B\cos(\pi/2)=0 B ′ = B cos ( π /2 ) = 0 A − C A-C A − C
分類 x 2 − 1 = 0 x^2-1=0 x 2 − 1 = 0 y − 1 = 0 y-1=0 y − 1 = 0 Δ = 0 \Delta=0 Δ = 0
解答 · 檢查 2 的答案 前者是兩條平行直線 x = ± 1 x=\pm1 x = ± 1 y = 1 y=1 y = 1
練習
用 θ = π / 4 \theta=\pi/4 θ = π /4 x y = 1 xy=1 x y = 1 x y = 0 xy=0 x y = 0
分類 x 2 + 4 y 2 − 2 x + 16 y + 17 = 0 x^2+4y^2-2x+16y+17=0 x 2 + 4 y 2 − 2 x + 16 y + 17 = 0 x 2 + 4 y 2 − 2 x + 16 y + 18 = 0 x^2+4y^2-2x+16y+18=0 x 2 + 4 y 2 − 2 x + 16 y + 18 = 0
求 x 2 + 4 y 2 = 4 x^2+4y^2=4 x 2 + 4 y 2 = 4 ( 0 , 1 ) (0,1) ( 0 , 1 ) ( 2 , 0 ) (2,0) ( 2 , 0 )
分類秩零方程 2 x − y + 3 = 0 2x-y+3=0 2 x − y + 3 = 0 0 = 0 0=0 0 = 0 1 = 0 1=0 1 = 0
引導解答
解答 · 解答 1 代入得 x y = ( X 2 − Y 2 ) / 2 xy=(X^2-Y^2)/2 x y = ( X 2 − Y 2 ) /2 x y = 1 xy=1 x y = 1 X 2 / 2 − Y 2 / 2 = 1 X^2/2-Y^2/2=1 X 2 /2 − Y 2 /2 = 1 2 \sqrt2 2 x y = 0 xy=0 x y = 0 X 2 − Y 2 = 0 X^2-Y^2=0 X 2 − Y 2 = 0
解答 · 解答 2 兩式左側分別化為 ( x − 1 ) 2 + 4 ( y + 2 ) 2 (x-1)^2+4(y+2)^2 ( x − 1 ) 2 + 4 ( y + 2 ) 2 ( x − 1 ) 2 + 4 ( y + 2 ) 2 + 1 (x-1)^2+4(y+2)^2+1 ( x − 1 ) 2 + 4 ( y + 2 ) 2 + 1 ( 1 , − 2 ) (1,-2) ( 1 , − 2 ) Δ = − 16 \Delta=-16 Δ = − 16
解答 · 解答 3 梯度為 ( 2 x , 8 y ) (2x,8y) ( 2 x , 8 y ) ( 0 , 1 ) (0,1) ( 0 , 1 ) ( 0 , 8 ) (0,8) ( 0 , 8 ) ( 2 , 0 ) (2,0) ( 2 , 0 ) ( 4 , 0 ) (4,0) ( 4 , 0 ) y = 1 y=1 y = 1 x = 2 x=2 x = 2 x x x y y y
解答 · 解答 4 軌跡依次為直線、整個平面、空集。只有第一個多項式具有非零梯度 ( 2 , − 1 ) (2,-1) ( 2 , − 1 ) 1 1 1