Question

Assinale a alternativa que contém as coordenadas cartesianas do foco desta hipérbole com a excentricidade e = $\sqrt{5}$ e vértices A1 (2,0) e A2 (-2,0) :

$a) (2 \sqrt{5} ,2 \sqrt{5} )$
$b) (2 \sqrt{5} ,0 )$
$c) (0,2 \sqrt{5} )$
$d) (-2 \sqrt{5},2 \sqrt{5} )$
$e) (-2 \sqrt{5},-2 \sqrt{5} )$

Answer 1

Boa noite Roger!

Solução!

$V1(-2,0)\\\\\\ V2(2,0)\\\\\ e= \sqrt{5}$

$Exentricidade\\\\\\\ e= \dfrac{c}{a}$


$Distancia ~~de~~V1~~a~~V2=4\\\\\\\ 2a=4\\\\\\ a= \dfrac{4}{2}\\\\\\ a=2\\\\\\\ Logo!\\\\\\ \dfrac{c}{a}= \sqrt{5}\\\\\\\ \dfrac{c}{2}= \sqrt{5}\\\\\\\ \boxed{c=2 \sqrt{5}}$

Como as distâncias focais  são simétricas temos!


$\boxed{-2 \sqrt{5},2 \sqrt{5}}\\\\\\\\\\\ \boxed{Resposta:~~Alternativa~~D}$


Boa noite!

Bons estudos!