[tex3]
x^2-y^2=1\iff y^2=x^2-1\quad\text{definido para }|x|\geqslant1\\[12pt]
1)\text{ Se }y\geq0\\[12pt]
y=\sqrt{x^2-1}\implies \dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{2x}{2\sqrt{x^2-1}}=\dfrac{x}{\sqrt{x^2-1}}=\dfrac{x}{y}\\
t\text { tangente de }\mathcal{C}\text{ em }(x_0,y_0):\\[12pt]
\dfrac{y-y_0}{x-x_0}=\dfrac{\mathrm{d}y}{\mathrm{d}x}(x_0)\implies y=\dfrac{x_0}{y_0}x-\dfrac{x_0^2}{y_0}+y_0=\dfrac{x_0}{y_0}x+\dfrac{y_0^2-x_0^2}{y_0}=\dfrac{x_0}{y_0}x-\dfrac{1}{y_0}\\[24pt]
n\text{ reta perpendicular a }t\text{ passanda por }(0,0):\\[12pt]
\dfrac{y-0}{x-0}=-\dfrac{y_0}{x_0}\implies y=-\dfrac{y_0}{x_0}x\\[24pt]
\text{e }I(x,y)\text{ intersecção de }n\text{ e }t\text{ tal que:}\\[12pt]
\begin{align}
\dfrac{x_0}{y_0}x-\dfrac{1}{y_0}=-\dfrac{y_0}{x_0}x\iff x=\dfrac{x_0}{2x_0^2-1}\\
\end{align}\\
\text{e então }y=-\dfrac{y_0}{x_0}\cdot\dfrac{x_0}{2x_0^2-1}=-\dfrac{y_0}{2x_0^2-1}\\
\text{e portanto temos }x^2+y^2=\dfrac{1}{2x_0^2-1}\\[24pt]
\begin{array}{}
x^2=\dfrac{x_0^2}{(2x_0^2-1)^2}&\implies 4x^2x_0^4-(4x^2+1)x_0^2+x^2=0\\
&\implies x_0^2=\dfrac{4x^2+1+\sqrt{8x^2+1}}{8x^2}\\
&\implies 2x_0^2-1=\dfrac{1+\sqrt{8x^2+1}}{4x^2}\\
&\implies x^2+y^2=\dfrac{4x^2}{1+\sqrt{8x^2+1}}=\dfrac{4x^2(\sqrt{8x^2+1}-1)}{8x^2}=\dfrac{\sqrt{8x^2+1}-1}{2}\\
&\implies (2x^2+2y^2+1)^2=8x^2+1\\
&\implies 4x^4+4y^4-4x^2+4y^2+8x^2y^2=0\\
&\implies x^4+y^4+2x^2y^2-x^2+y^2=0\\
&\implies (x^2+y^2)^2=x^2-y^2\\
\end{array}\\[96pt]
2)\text{ Se }y\leqslant0\text{ obtemos o mesmo resultado}\\[24pt]
\hspace{2cm}\boxed{\hspace{1cm}\\\hspace{0.5cm}(x^2+y^2)^2=x^2-y^2\hspace{0.5cm}\\\hspace{0.5cm}}
[/tex3]
- geogebra-export (16).png (60.34 KiB) Exibido 641 vezes