高専の数学解説

第15章 15.24

ax2+2bxy+cy2=1ax^2 + 2bxy + cy^2 = 1

(x′y′)=(cos⁡θ−sin⁡θsin⁡θcos⁡θ)(xy)\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}

から与式は

a(x′cos⁡θ−y′sin⁡θ)2+2b(x′cos⁡θ−y′sin⁡θ)(x′sin⁡θ+y′cos⁡θ)+c(x′sin⁡θ+y′cos⁡θ)2=1a(x'\cos\theta - y'\sin\theta)^2 + 2b(x'\cos\theta - y'\sin\theta)(x'\sin\theta + y'\cos\theta) + c(x'\sin\theta + y'\cos\theta)^2 = 1

x′2(acos⁡2θ+2bcos⁡θ⋅sin⁡θ+csin⁡θ)+y′2(a2sin⁡2θ−2bcos⁡θsin⁡θ+ccos⁡2θ)x'^2(a\cos^2\theta + 2b\cos\theta \cdot \sin\theta + c\sin\theta) + y'^2(a^2\sin^2\theta - 2b\cos\theta\sin\theta + c\cos^2\theta) +xy(−2acos⁡θsin⁡θ+2b(cos⁡2θ−sin⁡2θ)+2ccos⁡θsin⁡θ)=1+ xy(-2a\cos\theta\sin\theta + 2b(\cos^2\theta - \sin^2\theta) + 2c\cos\theta\sin\theta) = 1

x軸とy軸に一致させるには

−2acos⁡θsin⁡θ+2b(cos⁡2θ−sin⁡2θ)+2ccos⁡θsin⁡θ=0-2a\cos\theta\sin\theta + 2b(\cos^2\theta - \sin^2\theta) + 2c\cos\theta\sin\theta = 0

2(c−a)cos⁡θsin⁡θ+2b(cos⁡2θ−sin⁡2θ)=02(c-a)\cos\theta\sin\theta + 2b(\cos^2\theta - \sin^2\theta) = 0

(c−a)sin⁡2θ+2bcos⁡2θ=0(c-a)\sin2\theta + 2b\cos2\theta = 0

sin⁡2θcos⁡2θ=2ba−c\frac{\sin2\theta}{\cos2\theta} = \frac{2b}{a-c}

tan⁡2θ=2ba−c\tan2\theta = \frac{2b}{a-c}

解説: r31bn1z