fracx^236+fracy^29 = 1

常见问题解答

• Encontre o centro da elipse \(\frac{x^2}{36}+\frac{y^2}{9} = 1\)

  

  $$\left(0,0\right)$$

• Encontre os focos da elipse \(\frac{x^2}{36}+\frac{y^2}{9} = 1\)

  

  $$F_1=\left(-3\sqrt{3},0\right),F_2=\left(3\sqrt{3},0\right)$$

• Encontre os vértices da elipse \(\frac{x^2}{36}+\frac{y^2}{9} = 1\)

  

  $$V_1=\left(0,3\right),V_2=\left(0,-3\right),V_3=\left(6,0\right),V_4=\left(-6,0\right)$$

• Encontre a excentricidade da elipse \(\frac{x^2}{36}+\frac{y^2}{9} = 1\)

  

  $$e=\frac{\sqrt{3}}{2}$$

• $$\text{Resolva para }x$$ \(\frac{x^2}{36}+\frac{y^2}{9} = 1\)

  

  $$\begin{array}{rl}& x=2\sqrt{9-y^2}\\ & x=-2\sqrt{9-y^2}\end{array}$$

• $$\text{Resolva para }y$$ \(\frac{x^2}{36}+\frac{y^2}{9} = 1\)

  

  $$\begin{array}{rl}& y=\frac{\sqrt{36-x^2}}{2}\\ & y=-\frac{\sqrt{36-x^2}}{2}\end{array}$$

• Teste de simetria sobre a origem \(\frac{x^2}{36}+\frac{y^2}{9} = 1\)

  

  $$\text{关于原点对称}$$

• Testando a simetria em torno do eixo x \(\frac{x^2}{36}+\frac{y^2}{9} = 1\)

  

  $$\text{关于x轴对称}$$

• Testando a simetria sobre o eixo y \(\frac{x^2}{36}+\frac{y^2}{9} = 1\)

  

  $$\text{关于y轴对称}$$

• $$\text{Encontre a derivada em relac¸a˜o a }x$$ \(\frac{x^2}{36}+\frac{y^2}{9} = 1\)

  

  $$\frac{dy}{dx}=-\frac{x}{4y}$$

• $$\text{Encontre a derivada em relac¸a˜o a }y$$ \(\frac{x^2}{36}+\frac{y^2}{9} = 1\)

  

  $$\frac{dx}{dy}=-\frac{4y}{x}$$

• $$\text{Encontre a segunda derivada em relac¸a˜o a }x$$ \(\frac{x^2}{36}+\frac{y^2}{9} = 1\)

  

  $$\frac{d^{2}y}{d{x}^2}=-\frac{4y^2+x^2}{16y^3}$$

• $$\text{Encontre a segunda derivada em relac¸a˜o a }y$$ \(\frac{x^2}{36}+\frac{y^2}{9} = 1\)

  

  $$\frac{d^{2}x}{d{y}^2}=-\frac{4x^2+16y^2}{x^3}$$

• Reescrever na forma polar \(\frac{x^2}{36}+\frac{y^2}{9} = 1\)

  

  $$\begin{array}{rl}& r=\frac{6\sqrt{1+3{\sin }^2\left(\theta \right)}}{1+3{\sin }^2\left(\theta \right)}\\ & r=-\frac{6\sqrt{1+3{\sin }^2\left(\theta \right)}}{1+3{\sin }^2\left(\theta \right)}\end{array}$$