Question

Encontre o centro e os focos da cônica onde seus pontos satisfazem a seguinte condição: A distância ao ponto (2,2) é duas vezes a distância à reta y−3=0

Answer 1

Centro e Focos da cônica cuja a distância ao ponto (2,2) é igual a 2 vezes a distância à reta y-3 = 0.

Sejam os pontos (x,y). Que satisfazem :

$\displaystyle \sqrt{(\text x-2)^2+(\text y-2)^2} = 2.\frac{|0.\text x + 1.\text y - 3| }{\sqrt{0^2+1^2}}$

$\displaystyle \sqrt{(\text x-2)^2+(\text y-2)^2} = 2.|\text y - 3 |$

Elevando ao quadrados dos dois lados :

$(\text x -2)^2 + (\text y-2)^2 = 4.(\text y-3)^2$

$(\text x -2)^2 + \text y^2 - 4\text y + 4 = 4.(\text y^2 - 6\text y +9)$

$(\text x -2)^2 = 4.\text y^2 - 24\text y +36 - \text y^2 +4\text y - 4$

$(\text x -2)^2 = 3\text y^2 - 20\text y +32$

dividindo os dois lados por 3 :

$\displaystyle \frac{(\text x -2)^2}{3} = \text y^2 - \frac{20\text y}{3} +\frac{32}{3}$

completando quadrados somando k² dos dois lados  :

$\displaystyle \frac{(\text x -2)^2}{3} +\text k^2 = \text y^2 - \frac{20\text y}{3} +\frac{32}{3} + \text k^2$

analisando o produto notável de y com k :

$(\text y - \text k)^2 = \text y^2 -2.\text y.\text k + \text k^2$

Então basta que :

$\displaystyle 2\text y.\text k = \frac{20\text y }{3} \to \boxed{\text k = \frac{10}{3} }$

então fica assim :

$\displaystyle \frac{(\text x -2)^2}{3} +\frac{100}{9} = \text y^2 - \frac{20\text y}{3} + \frac{100}{9} +\frac{32}{3}$

$\displaystyle \frac{(\text x -2)^2}{3} +\frac{100}{9} = (\text y - \frac{10}{3})^2 +\frac{32}{3}$

$\displaystyle \frac{(\text x -2)^2}{3} - (\text y - \frac{10}{3})^2 = \frac{32}{3} - \frac{100}{9}$

$\displaystyle \frac{(\text x -2)^2}{3} - (\text y - \frac{10}{3})^2 = \frac{96}{9} - \frac{100}{9}$

$\displaystyle \frac{(\text x -2)^2}{3} - (\text y - \frac{10}{3})^2 = \frac{-4}{9}$

dividindo os dois lados por -4/9 :

$\displaystyle -\frac{(\text x -2)^2}{\displaystyle \frac{3}{\frac{4}{9}}} + \frac{(\text y - \frac{10}{3})^2}{\displaystyle \frac{4}{9}} = 1$

$\displaystyle -\frac{(\text x -2)^2}{\displaystyle \frac{27}{4}} + \frac{(\text y - \frac{10}{3})^2}{\displaystyle \frac{4}{9}} = 1$

$\displaystyle \frac{(\text y - \frac{10}{3})^2}{\displaystyle \frac{4}{9}} -\frac{(\text x -2)^2}{\displaystyle \frac{27}{4}} = 1$

Isso é uma hipérbole que está na horizontal.

portanto :

$\displaystyle \text a^2 = \frac{4}{9} \to \text a = \sqrt{\frac{4}{9}} \to \boxed{\text a = \frac{2}{3}}$  

$\displaystyle \text b^2 =\frac{27}{4} \to \text b = \sqrt{\frac{27}{4}} \to \boxed{\text b = \frac{3\sqrt{3}}{2}}$

$\displaystyle \text c^2 = \text a^2+\text b^2 \to c^2 = \frac{4}{9} + \frac{27}{4} \to \boxed{\text c= \frac{\sqrt{259}}{6}}$

$\huge\boxed{ \displaystyle \text{Centro}: (2,\frac{10}{3})} \checkmark$

Focos :

$\displaystyle \text F_1 : (2, \frac{10}{3} - \text c) \to \text F_1 : (2, \frac{10}{3} - \frac{\sqrt{259}}{6}) \to \text F_1 : ( 2, \frac{20-\sqrt{259}}{6} )$

$\displaystyle \text F_2 : (2, \frac{10}{3} + \text c) \to \text F_2 : (2, \frac{10}{3} + \frac{\sqrt{259}}{6}) \to \text F_2 : ( 2, \frac{20+\sqrt{259}}{6} )$  

$\huge\boxed{\displaystyle \text F_1 : ( 2, \frac{20-\sqrt{259}}{6} ) \ ; \text F_2 : (2,\frac{20+\sqrt{259}}{6}) }\checkmark$