236. Etudiez la nature d’une conique (1/2)

x^2+\sqrt{3}xy+x-2=0.

\left\{
\begin{align*}
\vv{u} &= \cos \theta \vv{i} + \sin \theta \vv{j}\\
\vv{v} &= -\sin \theta \vv{i} + \cos \theta \vv{j}.
\end{align*}\right.

\begin{align*}
\vv{OM} &= X \vv{u} + Y \vv{v}\\
&=X( \cos \theta \vv{i} + \sin \theta \vv{j})+Y(-\sin \theta \vv{i} + \cos \theta \vv{j})\\
&=(X\cos \theta - Y\sin \theta)\vv{i}+(X\sin \theta+Y\cos \theta)\vv{j}.
\end{align*}

\boxed{\begin{align*}
x&=X\cos \theta - Y\sin \theta \\
y&=X\sin \theta+Y\cos \theta.
\end{align*}}

\begin{align*}
X&=x\cos \theta + y\sin \theta \\
Y&=-x\sin \theta+y\cos \theta.
\end{align*}

\begin{align*}
x\cos \theta + y\sin\theta &= (X\cos \theta- Y\sin\theta)\cos \theta + (X\sin\theta+Y\cos\theta)\sin\theta\\
&= X\cos^2 \theta+ X\sin^2\theta\\
&=X(\cos^2 \theta+ \sin^2\theta)\\
&=X.
\end{align*}

\begin{align*}
-x\sin \theta + y\cos\theta &= (-X\cos\theta +Y\sin\theta)\sin\theta + (X\sin \theta+Y\cos \theta)\cos\theta\\
&= Y\sin^2\theta + Y\cos^2\theta\\
&=Y(\sin^2\theta+\cos^2\theta)\\
&=Y.
\end{align*}

\begin{align*}
x^2+\sqrt{3}xy+x-2=0 &\Longleftrightarrow (X\cos \theta - Y\sin \theta)^2+ \sqrt{3}(X\cos \theta - Y\sin \theta)(X\sin \theta+Y\cos \theta)+X\cos \theta - Y\sin \theta - 2 = 0\\
&\Longleftrightarrow X^2\cos^2\theta+Y^2\sin^2\theta-2XY\sin\theta \cos\theta+ \sqrt{3}(X^2\sin\theta \cos\theta-Y^2\sin\theta\cos\theta + XY(\cos^2\theta - \sin^2\theta))+X\cos \theta - Y\sin \theta - 2 = 0\\
&\Longleftrightarrow (\cos^2\theta+\sqrt{3}\sin\theta\cos\theta)X^2+(\sin^2\theta-\sqrt{3}\sin\theta\cos\theta)Y^2+(\sqrt{3}(\cos^2\theta - \sin^2\theta)-2\sin\theta\cos\theta)XY+X\cos \theta - Y\sin \theta - 2 = 0.
\end{align*}

\begin{align*}
\cos^2\theta &=\frac{\cos^2\theta}{1}\\
&=\frac{\cos^2\theta}{\sin^2\theta + \cos^2\theta}\\
&=\frac{1}{u^2 + 1} 
\end{align*} 

\begin{align*}
\sin^2\theta &=1 - \cos^2\theta\\
&=1-\frac{1}{1+u^2}\\
&=\frac{u^2}{1+u^2}.
\end{align*} 

\begin{align*}
\cos^2\theta - \sin^2\theta &=\frac{1-u^2}{1+u^2}.
\end{align*}

\begin{align*}
\sin\theta\cos\theta&=\frac{\sin\theta}{\cos\theta}\cos^2\theta\\
&= u\cos^2\theta\\
&=\frac{u}{1+u^2}.
\end{align*}

\begin{align*}
x^2+\sqrt{3}xy+x-2=0 
&\Longleftrightarrow \left(\frac{1}{1+u^2}+\frac{\sqrt{3}u}{1+u^2}\right)X^2+\left(\frac{u^2}{1+u^2}-\frac{\sqrt{3}u}{1+u^2}\right)Y^2+\left(\sqrt{3}\frac{1-u^2}{1+u^2}-2\frac{u}{1+u^2}\right)XY+X\cos \theta - Y\sin \theta - 2 = 0\\
&\Longleftrightarrow (1+\sqrt{3}u)X^2+(u^2-\sqrt{3}u)Y^2+(\sqrt{3}(1-u^2)-2u)XY+X\cos \theta (1+u^2)- Y\sin \theta (1+u^2)- 2(1+u^2) = 0.
\end{align*}

\begin{align*}
\sqrt{3}(1-u^2)-2u &=0\\
3(1-u^2)-2\sqrt{3}u &=0\\
3-3u^2-2\sqrt{3}u &=0\\
3u^2+2\sqrt{3}u -3&=0\\
(\sqrt{3}u+1)^2-4&=0\\
(\sqrt{3}u+1)^2-2^2&=0\\
(\sqrt{3}u-1)(\sqrt{3}u+3)&=0\\
(3u-\sqrt{3})(3u+3\sqrt{3})&=0\\
(3u-\sqrt{3})(u+\sqrt{3})&=0\\
\end{align*}

\begin{align*}
\cos\theta &= \frac{1}{\sqrt{1+u^2}}\\
 &= \frac{1}{\sqrt{1+\frac{1}{3}}}\\
&= \frac{1}{\sqrt{\frac{4}{3}}}\\
&= \frac{\sqrt{3}}{2}.
\end{align*}

\begin{align*}
\sin\theta &=u\cos\theta\\
 &=\frac{\sqrt{3}}{3}\times \frac{\sqrt{3}}{2}\\
&= \frac{1}{2}.
\end{align*}

\begin{align*}
1+\sqrt{3}u &=1+\sqrt{3} \times \frac{\sqrt{3}}{3}\\
&=2.
\end{align*}

\begin{align*}
u^2-\sqrt{3}u &= \frac{1}{3}- 1\\
&=\frac{-2}{3}.
\end{align*}

\begin{align*}
1+u^2 &= 1+\frac{1}{3}\times 1\\
&=\frac{4}{3.}
\end{align*}

\begin{align*}
x^2+\sqrt{3}xy+x-2=0 
&\Longleftrightarrow 2X^2-\frac{2}{3}Y^2+\frac{\sqrt{3}}{2}\times \frac{4}{3}X- \frac{1}{2}\times \frac{4}{3}Y-\frac{8}{3} = 0\\
&\Longleftrightarrow 2X^2-\frac{2}{3}Y^2+\frac{2\sqrt{3}}{3}X- \frac{2}{3}Y-\frac{8}{3} = 0\\
&\Longleftrightarrow 6X^2-2Y^2+2\sqrt{3}X- 2Y-8 = 0\\
&\Longleftrightarrow 3X^2-Y^2+\sqrt{3}X- Y-4 = 0.
\end{align*}

\boxed{x^2+\sqrt{3}xy+x-2=0   \Longleftrightarrow 3X^2-Y^2+\sqrt{3}X- Y-4 = 0.}