二直線の交角の二等分線を求める

(問題)
　\(ax^2+2hxy+by^2=0\)　ヲ以テ表ハセレタル二直線ノ交角ヲ二等分スル直線ノ方程式ヲ求ム。
(　\(ax^2+2hxy+by^2=0\)で表される二直線の交角を二等分する直線の方程式を求めよ。）

\begin{align*}ax^2+2hxy=0 　\Longleftrightarrow 　x(ax+2hy)=0\end{align*}.

\begin{align*} \left| X \right| = \frac{\left| aX + 2hY \right|}{\sqrt{a^2 + (2h)^2}} &\Longleftrightarrow (a^2 + 4h^2)X^2 = a^2X^2 + 4ahXY + 4h^2Y^2\\\\ &\Longleftrightarrow hY^2 + ahXY – hX^2 = 0\\\\ &\Longleftrightarrow Y=\frac{-a±\sqrt{a^{2}+4h^{2}}}{2h}X. \end{align*}

\begin{align*} bt^2 + 2ht + \alpha = 0 \end{align*}

\begin{align*} ax^2+2hxy + by^2 =0 \Longleftrightarrow (y-\alpha x)(y-\beta x)=0. \end{align*}

\begin{align*} &\frac{|\alpha X − Y|}{\sqrt{\alpha ^{2} + 1}} = \frac{|−\beta X + Y|}{ \sqrt{\beta ^{2} + 1}}\\ \Longleftrightarrow &(\beta ^{2} + 1)(Y^{2} − 2\alpha XY + \alpha ^{2}X^{2}) = (\alpha ^{2} + 1)(Y^{2} − 2\beta XY + \beta^{2}X^{2})\\ \Longleftrightarrow &(\beta ^{2} − \alpha ^{2})Y^{2} + 2(−\alpha + \beta – \alpha \beta ^{2} + \alpha ^{2} \beta)XY + (\alpha ^{2} − \beta ^{2})X^{2} = 0\\ \Longleftrightarrow &(\alpha + \beta)Y^{2} + 2(1 − \alpha \beta)XY − (α + β)X^{2} = 0.　 \cdots ① \end{align*}

\begin{align*} ① &\Longleftrightarrow -\frac{2h}{b}Y^{2} + 2\left(1-\frac{a}{b}\right)XY + \frac{2h}{b}X^{2} = 0 \\ &\Longleftrightarrow hY^{2} + (a – b)XY – hX^{2} = 0 \\ &\Longleftrightarrow Y = \frac{-a + b \pm \sqrt{(a – b)^{2} + 4h^{2}}}{2h} X \end{align*}

\begin{eqnarray} \left\{ \begin{array}{l} x=0 , y=0 &(h=0 , ab\lt 0) \\ y=\dfrac{-a + b \pm \sqrt{(a – b)^{2} + 4h^{2}}}{2h} x &(h\neq 0 , h^{2} \gt ab). \end{array} \right. \end{eqnarray}