Question

onsidere a equação da elipse 16 x + 4 y -256=0 obtenha os vertices do eixo maior e sua excentricidade, essa aqui cheguei a1 = 0,-8 e a2 = 0,8 e excentricidade ta me matando nao consigo chegar no resultado pode me ajudar ??

Answer 1

Boa tarde!

Solução!

Equação da elipse!


$\dfrac{ x^{2} }{a^{2} }+ \dfrac{y^{2} }{b^{2} }=1\\\\\\\\ a\ \textgreater \ b~~O~~eixo~~maior~~esta~~contido~~em~~x\\\\\\\ b\ \textgreater \ a~~O~~eixo~~maior~~esta~~contido~~em~~y$


$Excentricidade~~e\´~~dada ~~pela~~formula.$

$e= \frac{c}{a}$

$16 x^{2} +4y^{2}-256=0\\\\\\\ 16 x^{2} +4y^{2}=256\\\\\\\ \dfrac{16 x^{2} }{256}+ \dfrac{4y^{2} }{256}= \dfrac{256}{256}\\\\\\\\ \dfrac{ x^{2} }{16}+ \dfrac{y^{2} }{64}=1\\\\\\\\ Veja!\\\\\ b\ \textgreater \ a\\\\\\\ Vertice= \sqrt{64}\\\\ Vertice=\pm8\\\\\\ V1(0,8)\\\\\\\ V2(0,-8)$


$b^{2}=a^{2}+c^{2}\\\\\\ 8^{2}=4^{2}+ c^{2}\\\\\\ 64=16+ c^{2}\\\\\\ 48=c^{2}\\\\\\ c= \sqrt{48} \\\\\\ c=4 \sqrt{3}\\\\\\\\\ e= \frac{4 \sqrt{3} }{4} \\\\\\\ \boxed{e= \sqrt{3} }$


Boa noite!
Bons estudos!