こんにちは。知っておくと意外と便利な公式を提示します。よければ使ってください。検算などには役立つと思います。
放物線とグラフ上の2点A, Bにおける接線
,
が点Pで交わる。このとき, 交点Pの座標を求めてみることにする。
すべての放物線は相似の関係にあるので,
![Rendered by QuickLaTeX.com y=ax^2+bx+c](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-642a359ee1a5612768e5b8ebdaea0a40_l3.png)
![Rendered by QuickLaTeX.com y=ax^2](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-3d90f33a3d8d1a392fac89732ce270f4_l3.png)
放物線の式を
![Rendered by QuickLaTeX.com y=ax^2](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-3d90f33a3d8d1a392fac89732ce270f4_l3.png)
![Rendered by QuickLaTeX.com y'=2ax](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-901c176f2c778c21fded7bf73cb19111_l3.png)
![Rendered by QuickLaTeX.com \mathrm{A}(\alpha, a\alpha^2)](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-02661a1206e130b1b467ae1dd5b4cf8d_l3.png)
![Rendered by QuickLaTeX.com \mathrm{B}(\beta, a\beta^2)](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-b468992c3090a36fcb377363209b102e_l3.png)
![Rendered by QuickLaTeX.com \alpha<\beta](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-ea58315fc8cc8260f3d2c45e1bfba225_l3.png)
![Rendered by QuickLaTeX.com \ell_a, \ell_b](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-1db46522e7b4dea69e1c3a7c0b44e9e9_l3.png)
![Rendered by QuickLaTeX.com \ell_a : y=2a\alpha(x-\alpha)+a\alpha^2\to y=2a\alpha x-a\alpha^2\cdots\maru1](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-fb8dc7a7b7e94a2b50a2c8aa24d1f4fc_l3.png)
![Rendered by QuickLaTeX.com \ell_b : y=2a\beta(x-\beta)+a\beta^2\to y=2a\beta x-a\beta^2\cdots\maru2](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-adc6c7231877ba6dc7df7ccd8fa77ce4_l3.png)
よって交点Pの
![Rendered by QuickLaTeX.com x](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-5f16b0dcec027c9742e11d99170299a8_l3.png)
![Rendered by QuickLaTeX.com 2a\alpha x-a\alpha^2=2\beta x-a\beta^2](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-f8294b4265331b0580f4c4642746ccdf_l3.png)
![Rendered by QuickLaTeX.com 2a(\alpha-\beta)x=a(\alpha+\beta)(\alpha-\beta)](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-74cb6b1e0e4c877d36af22c27ea7ede6_l3.png)
![Rendered by QuickLaTeX.com x=\dfrac{\alpha+\beta}{2}\cdots\maru3](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-b7d9d8e2edf851868c83cf23fea8f259_l3.png)
![Rendered by QuickLaTeX.com \maru3](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-59e0023e63fc0665bd608c9e370731c9_l3.png)
![Rendered by QuickLaTeX.com \maru1](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-c8b8ec9c0d15342374d474f3407d687d_l3.png)
![Rendered by QuickLaTeX.com y=2a\alpha\cdot\dfrac{\alpha+\beta}{2}-a \alpha^2=a\alpha\beta](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-7de8424f0f782501ad629b8163a6191a_l3.png)
以上より, 点Pの座標は,
![Rendered by QuickLaTeX.com \mathrm{P}\left(\dfrac{\alpha+\beta}{2}, a\alpha\beta\right)](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-7e3977f51372548499512f79be10b69d_l3.png)
このように, 点Pの
![Rendered by QuickLaTeX.com x](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-5f16b0dcec027c9742e11d99170299a8_l3.png)
交点Pの座標がが2点A, Bの中点であることがわかった。ここで, 線分ABの中点をMとすると,
となる。
ここで, 線分PMの中点Nを求めてみると,
![Rendered by QuickLaTeX.com x](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-5f16b0dcec027c9742e11d99170299a8_l3.png)
![Rendered by QuickLaTeX.com \dfrac{\alpha+\beta}{2}](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-ef794262daa773921a211fe61af7242f_l3.png)
![Rendered by QuickLaTeX.com y](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-5d76ceac31cb52dd9eb4431a14c502dc_l3.png)
![Rendered by QuickLaTeX.com \left\{a\alpha\beta+\dfrac{a(\alpha^2+\beta^2)}{2}\right\}\times\dfrac12](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-35b7a8d5208a31e13d07650585d2ac22_l3.png)
![Rendered by QuickLaTeX.com \to\left\{\dfrac{a(\alpha^2+2\alpha\beta+\beta^2)}{4}\right\}](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-e36c75f8c487908e826af1049cac9234_l3.png)
![Rendered by QuickLaTeX.com \to a\left(\dfrac{\alpha+\beta}{2}\right)^2](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-41b4b2e79bfedc8370fd0a8de77b0e96_l3.png)
となり, Nの座標は,
![Rendered by QuickLaTeX.com \mathrm{N}\left(\dfrac{\alpha+\beta}{2}, a\left(\dfrac{\alpha+\beta}{2}\right)^2\right)](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-f76659faf1d72eb2c01ed1ffd2b6050d_l3.png)
である。この座標は, 放物線
![Rendered by QuickLaTeX.com y=ax^2](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-3d90f33a3d8d1a392fac89732ce270f4_l3.png)
線分PMの中点Nは放物線上の点になる。
ついでに, MN, NPの長さを求めてみると,
![Rendered by QuickLaTeX.com \mathrm{MN}=\mathrm{NP}=\dfrac12\mathrm{PM}](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-e7807910a67abb8eb007e8a74838d9c3_l3.png)
![Rendered by QuickLaTeX.com \dfrac12\mathrm{PM}=\dfrac12\left\{\dfrac{a(\alpha^2+\beta^2)}{2}-a\alpha\beta\right\}\to a\left(\dfrac{\alpha-\beta}{2}\right)^2](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-b57fe1fb695218dff384fbd774d216e9_l3.png)
よって,
![Rendered by QuickLaTeX.com \mathrm{MN}=\mathrm{NP}=a\left(\dfrac{\alpha-\beta}{2}\right)^2](https://mathtext.info/blog/wordpress/wp-content/ql-cache/quicklatex.com-4d5861f14692a947c6f139205b0e0f6b_l3.png)
となる。
最後にもう1つだけ。
グラフ上の点Nにおける接線の傾きは,
, Nの
座標が
であることから,
傾きは,
となる。
ここで, 直線ABの傾きを求めてみると,
,
であるから,
となり, 点Nにおける放物線の接線の傾きと, 直線ABの傾きは等しくなることがわかる。
つまり,
となる。