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数学検定 準1級受験メモ: 放物線の接線の方程式導出

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放物線とは

放物線の方程式は次の式で与えられる.
$$ y^2 = 4 p x$$

焦点は$F = (p, 0)$で与えられ,準線は$x = -p$ で与えられる.

図示すると以下のようになります.

放物線の接線の方程式の導出

まず $ y^2 = 4 p x$ の両辺を x で微分する.
$$ 2y \frac{dy}{dx} = 4p $$

よって,
$$ \frac{dy}{dx} = \frac{4p}{2y} = \frac{2p}{y} $$

さて,接線の方程式は$x$の一次式なので
$$ y - y_0 = m (x - x_0) $$
で与えられる.

ここで,$m$は接線の傾きであり,$(x_0, y_0)$は接線と放物線の接点の座標である.
接点にて接線の傾きは $m = \frac{dy}{dx} = \frac{2p}{y_0}$ である.
あくまで「接点での傾き」なので$\frac{2p}{y}$ではなく$\frac{2p}{y_0}$である.

この傾きを接線の方程式に代入すると
$$ y - y_0 = \frac{2p}{y_0} (x - x_0) $$
この式を展開していくと,

$$ y = \frac{2p}{y_0} (x - x_0) + y_0$$

$$ y_0 y = 2p (x - x_0) + {y_0}^2$$

$$ y_0 y = 2p (x - x_0) + 4 p x_0$$

$$ y_0 y = 2p (x + x_0)$$

以上,接線の方程式が求められました.

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