Question

Quais são as coordenadas dos pontos de interseção de uma parábola na forma y=x^2 com uma circunferência de raio 2 centrada na origem?

Answer 1

$\mathrm{f(x)=x^2\ \to\ par\'abola}\\\\ \mathrm{x^2+y^2=r^2\ \to\ y^2=2^2-x^2}\\ \mathrm{y=\sqrt{4-x^2}\ \to\ g(x)=\sqrt{4-x^2}\ \to\ circunfer\^encia}\\\\ \textrm{Abscissas de intersec\c{c}\~ao:}\\\\ \mathrm{f(x)=g(x)\ \to\ x^2=\sqrt{4-x^2}\ \to\ (x^2)^2=(\sqrt{4-x^2})^2}\\ \mathrm{x^4=4-x^2\ \to\ x^4+x^2-4=0}\\\\ \mathrm{Substituiremos\ x^2\ por\ u\ e\ x^4\ por\ u^2:}\\\\ \mathrm{u^2+u-4=0\ \to\ a=1\ \| \ b=1\ \| \ c=-4}$

$\textrm{A partir da f\'ormula quadr\'atica, obteremos:}\\\\ \mathrm{u=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\ \to\ u=\dfrac{-1\pm\sqrt{1^2-4.1.(-4)}}{2.1}}\\\\ \mathrm{u=\dfrac{-1\pm\sqrt{1+16}}{2}\ \to\ u=\dfrac{-1\pm\sqrt{17}}{2}}\\\\ \mathrm{u=\bigg\{\dfrac{-1+\sqrt{17}}{2},\dfrac{-1-\sqrt{17}}{2}\bigg\}}$

$\mathrm{Como\ u=x^2,\ teremos\ que:}\\\\ \mathrm{x^2=\dfrac{-1+\sqrt{17}}{2}\ \to\ x=\pm\sqrt{\dfrac{-1+\sqrt{17}}{2}}\ \to\ x\in\mathbb{R}}\\\\ \mathrm{x^2=\dfrac{-1-\sqrt{17}}{2}\ \to\ x=\pm\sqrt{\dfrac{-1-\sqrt{17}}{2}}\ \to\ x\in\mathbb{C}}\\\\ \textrm{Portanto:}\\\\ \mathrm{x=\bigg\{\sqrt{\dfrac{-1+\sqrt{17}}{2}},-\sqrt{\dfrac{-1+\sqrt{17}}{2}}\bigg\}}$

$\textrm{Ordenada de intersec\c{c}\~ao:}\\\\ \mathrm{f(x)=x^2\ \to\ f(x)=\bigg(\pm\sqrt{\dfrac{-1+\sqrt{17}}{2}}\bigg)^2}\\\\ \mathrm{f(x)=\dfrac{-1+\sqrt{17}}{2}\ \to\ y=\dfrac{-1+\sqrt{17}}{2}}\\\\ \textrm{Pontos de intersec\c{c}\~ao:}\\\\ \mathrm{P=(x,y)\ \to\ P=\bigg(\pm\sqrt{\dfrac{-1+\sqrt{17}}{2}}, \dfrac{-1+\sqrt{17}}{2}\bigg)}\\\\ \mathrm{P_1=\bigg(\sqrt{\dfrac{-1+\sqrt{17}}{2}},\dfrac{-1+\sqrt{17}}{2}\bigg)}\\\\ \mathrm{P_2=\bigg(-\sqrt{\dfrac{-1+\sqrt{17}}{2}},\dfrac{-1+\sqrt{17}}{2}\bigg)}$

Answer 2

Bom dia!

Usando a equação reduzida da circunferência temos:

$(x-x_c) ^{2} + (y-y_c) ^{2} = r ^{2}$

$(x-0) (x-0) + (y-0) (y-0)=2 ^{2}$

$x ^{2} + y^{2} =4$

Agora montamos um sistema:

$\left \{ {{y=} x^{2}(-1) \atop {2^{2}=y ^{2} }} \right.$

$-y+4=y ^{2}$

$y ^{2}+y-4=0$

$\Delta=1 ^{2} -4.1.-4$

$\Delta=17$

$y= \frac{-1+ \sqrt{\pm17}}{2}$ 

Agora encontramos o valor de x:

$y= x^{2}$

O valor desta raiz não pode ser negativa pois não existe raiz negativa no conjunto dos números reais.

$x=\pm\sqrt{\frac{-1+ \sqrt{17}}{2}$

$P1=(\sqrt{\frac{-1+ \sqrt{17}}{2}}, \frac{-1+ \sqrt{17}}{2})$

$P2=(-\sqrt{\frac{-1+ \sqrt{17}}{2}}, \frac{-1+ \sqrt{17}}{2})$

Espero ter ajudado.