HSC 擴充套件 1 和 2 數學/4 個單元/圓錐曲線

橢圓

橢圓的笛卡爾方程為 ${\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}=1$ 。求梯度時微分（使用隱式微分來簡化過程）

雙曲線的笛卡爾方程為 ${\frac {x^{2}}{a^{2}}}-{\frac {y^{2}}{b^{2}}}=1$ 。求梯度時微分（使用隱式微分來簡化過程）

{\begin{aligned}0&={\frac {2x}{a^{2}}}-{\frac {2y}{b^{2}}}{\frac {dy}{dx}}\\{\frac {dy}{dx}}{\frac {y}{b^{2}}}&={\frac {x}{a^{2}}}\\{\frac {dy}{dx}}&={\frac {b^{2}}{a^{2}}}\times {\frac {x}{y}}\end{aligned}}

然後我們可以將其代入我們的點斜式， $y-y_{1}=m(x-x_{1})$ ，使用點 $P(x_{1},y_{1})$

在

P

，

m={\frac {b^{2}}{a^{2}}}{\frac {x_{1}}{y_{1}}}

。

{\begin{aligned}y-y_{1}&={\frac {b^{2}}{a^{2}}}{\frac {x_{1}}{y_{1}}}(x-x_{1})\\yy_{1}-y_{1}^{2}&={\frac {b^{2}}{a^{2}}}x_{1}(x-x_{1})\\{\frac {yy_{1}}{b^{2}}}-{\frac {y_{1}^{2}}{b^{2}}}&={\frac {x_{1}}{a^{2}}}(x-x_{1})\\{\frac {yy_{1}}{b^{2}}}-{\frac {y_{1}^{2}}{b^{2}}}&={\frac {xx_{1}}{a^{2}}}-{\frac {x_{1}^{2}}{a^{2}}}\\{\frac {x_{1}^{2}}{a^{2}}}-{\frac {y_{1}^{2}}{b^{2}}}&={\frac {xx_{1}}{a^{2}}}-{\frac {yy_{1}}{b^{2}}}\\\end{aligned}}

但我們知道，根據雙曲線的定義，

{\frac {x_{1}^{2}}{a^{2}}}-{\frac {y_{1}^{2}}{b^{2}}}=1

，所以

{\frac {xx_{1}}{a^{2}}}-{\frac {yy_{1}}{b^{2}}}=1

法線的斜率由 $-{\frac {dx}{dy}}$ 給出，即 $-{\frac {a^{2}}{b^{2}}}\times {\frac {y}{x}}$ 。求出方程，

{\begin{aligned}y-y_{1}&=-{\frac {a^{2}}{b^{2}}}{\frac {y_{1}}{x_{1}}}(x-x_{1})\\yb^{2}-y_{1}b^{2}&=-a^{2}{\frac {y_{1}}{x_{1}}}(x-x_{1})\\{\frac {b^{2}y}{y_{1}}}-b^{2}&=-{\frac {a^{2}}{x_{1}}}(x-x_{1})\\{\frac {b^{2}y}{y_{1}}}-b^{2}&=-{\frac {a^{2}x}{x_{1}}}+a^{2}\\{\frac {a^{2}x}{x_{1}}}+{\frac {b^{2}y}{y_{1}}}&=a^{2}+b^{2}\end{aligned}}