N_ha

接線の方程式

数学
高校数学

作成日 2022年6月26日日曜日

更新日 2025年3月13日木曜日

放物線y2=4px⋯(∗)y^2=4px\cdots\left(*\right)上の点P(x1,y1)(≠(0,0))P\left(x_1, y_1\right)\left(\neq\left(0,0\right)\right)における接線の方程式

(∗)\left(*\right)の両辺をxxで微分して

ddxy2=ddx4pxdydxddyy2=4pdydx2y=4pdydx=4p2y=2py\begin{aligned} \frac{d}{dx}y^2&=\frac{d}{dx}4px\\ \frac{dy}{dx}\frac{d}{dy}y^2&=4p\\ \frac{dy}{dx}2y&=4p\\ \frac{dy}{dx}&=\frac{4p}{2y}\\&=\frac{2p}{y} \end{aligned}

よって、点PPにおける(∗)\left(*\right)の接線の方程式は

y−y1=2py1(x−x1)y1(y−y1)=2p(x−x1)y1y−y12=2px−2px1y1y=2px−2px1+y12=2px−2px1+4px1(∵y12=4px1)=2p(x+x1)\begin{aligned} y-y_1&=\frac{2p}{y_1}\left(x-x_1\right)\\ y_1\left(y-y_1\right)&=2p\left(x-x_1\right)\\ y_1y-y_1^2&=2px-2px_1\\ y_1y&=2px-2px_1+y_1^2\\&=2px-2px_1+4px_1\left(\because y_1^2=4px_1\right)\\&=2p\left(x+x_1\right) \end{aligned}

円x2+y2=r2⋯(∗)x^2+y^2=r^2\cdots\left(*\right)上の点P(x1,y1)(≠(−r,0),(r,0))P\left(x_1, y_1\right)\left(\neq\left(-r,0\right),\left(r,0\right)\right)における接線の方程式

(∗)\left(*\right)の両辺をxxで微分して

ddx(x2+y2)=ddxr2ddxx2+ddxy2=02x+dydxddyy2=0dydx2y=−2xdydx=−2x2y=−xy\begin{aligned} \frac{d}{dx}\left(x^2+y^2\right)&=\frac{d}{dx}r^2\\ \frac{d}{dx}x^2+\frac{d}{dx}y^2&=0\\ 2x+\frac{dy}{dx}\frac{d}{dy}y^2&=0\\ \frac{dy}{dx}2y&=-2x\\ \frac{dy}{dx}&=\frac{-2x}{2y}\\&=-\frac{x}{y} \end{aligned}

よって、点PPにおける(∗)\left(*\right)の接線の方程式は

y−y1=−x1y1(x−x1)y1(y−y1)=−x1(x−x1)y1y−y12=−x1x+x12x1x+y1y=x12+y12=r2(∵x12+y12=r2)\begin{aligned} y-y_1&=-\frac{x_1}{y_1}\left(x-x_1\right)\\ y_1\left(y-y_1\right)&=-x_1\left(x-x_1\right)\\ y_1y-y_1^2&=-x_1x+x_1^2\\ x_1x+y_1y&=x_1^2+y_1^2\\&=r^2\left(\because x_1^2+y_1^2=r^2\right) \end{aligned}

楕円x2a2+y2b2=1⋯(∗)\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\cdots\left(*\right)上の点P(x1,y1)(≠(−a,0),(a,0))P\left(x_1, y_1\right)\left(\neq\left(-a,0\right),\left(a,0\right)\right)における接線の方程式

(∗)\left(*\right)の両辺をxxで微分して

ddx(x2a2+y2b2)=ddx1ddxx2a2+ddxy2b2=02xa2+dydxddyy2b2=0dydx2yb2=−2xa2dydx=−2xa2⋅b22y=−b2xa2y\begin{aligned} \frac{d}{dx}\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)&=\frac{d}{dx}1\\ \frac{d}{dx}\frac{x^2}{a^2}+\frac{d}{dx}\frac{y^2}{b^2}&=0\\ \frac{2x}{a^2}+\frac{dy}{dx}\frac{d}{dy}\frac{y^2}{b^2}&=0\\ \frac{dy}{dx}\frac{2y}{b^2}&=-\frac{2x}{a^2}\\ \frac{dy}{dx}&=-\frac{2x}{a^2}\cdot\frac{b^2}{2y}\\&=-\frac{b^2x}{a^2y} \end{aligned}

よって、点PPにおける(∗)\left(*\right)の接線の方程式は

y−y1=−b2x1a2y1(x−x1)a2y1(y−y1)=−b2x1(x−x1)a2y1y−a2y12=−b2x1x+b2x12b2x1x+a2y1y=b2x12+a2y12x1xa2+y1yb2=x12a2+y12b2=1(∵x12a2+y12b2=1)\begin{aligned} y-y_1&=-\frac{b^2x_1}{a^2y_1}\left(x-x_1\right)\\ a^2y_1\left(y-y_1\right)&=-b^2x_1\left(x-x_1\right)\\ a^2y_1y-a^2y_1^2&=-b^2x_1x+b^2x_1^2\\ b^2x_1x+a^2y_1y&=b^2x_1^2+a^2y_1^2\\ \frac{x_1x}{a^2}+\frac{y_1y}{b^2}&=\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}\\&=1\left(\because \frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}=1\right) \end{aligned}

双曲線x2a2−y2b2=1⋯(∗)\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\cdots\left(*\right)上の点P(x1,y1)(≠(−a,0),(a,0))P\left(x_1, y_1\right)\left(\neq\left(-a,0\right),\left(a,0\right)\right)における接線の方程式

(∗)\left(*\right)の両辺をxxで微分して

ddx(x2a2−y2b2)=ddx1ddxx2a2−ddxy2b2=02xa2−dydxddyy2b2=0dydx2yb2=2xa2dydx=2xa2⋅b22ydydx=b2xa2y\begin{aligned} \frac{d}{dx}\left(\frac{x^2}{a^2}-\frac{y^2}{b^2}\right)&=\frac{d}{dx}1\\ \frac{d}{dx}\frac{x^2}{a^2}-\frac{d}{dx}\frac{y^2}{b^2}&=0\\ \frac{2x}{a^2}-\frac{dy}{dx}\frac{d}{dy}\frac{y^2}{b^2}&=0\\ \frac{dy}{dx}\frac{2y}{b^2}&=\frac{2x}{a^2}\\ \frac{dy}{dx}&=\frac{2x}{a^2}\cdot\frac{b^2}{2y}\\ \frac{dy}{dx}&=\frac{b^2x}{a^2y} \end{aligned}

よって、点PPにおける(∗)\left(*\right)の接線の方程式は

y−y1=b2x1a2y1(x−x1)a2y1(y−y1)=b2x1(x−x1)a2y1y−a2y12=b2x1x−b2x12b2x1x−a2y1y=b2x12−a2y12x1xa2−y1yb2=x12a2−y12b2=1(∵x12a2−y12b2=1)\begin{aligned} y-y_1&=\frac{b^2x_1}{a^2y_1}\left(x-x_1\right)\\ a^2y_1\left(y-y_1\right)&=b^2x_1\left(x-x_1\right)\\ a^2y_1y-a^2y_1^2&=b^2x_1x-b^2x_1^2\\ b^2x_1x-a^2y_1y&=b^2x_1^2-a^2y_1^2\\ \frac{x_1x}{a^2}-\frac{y_1y}{b^2}&=\frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}\\&=1\left(\because \frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}=1\right) \end{aligned}