Question

me ajudem f)289x²-17183=2(256y-289x-32y²)

Answer 1

Explicação passo-a-passo:

$\sf 289x^2-17183=2\cdot(256y-289x-32y^2)$

$\sf 289x^2-17183=512y-578x-64y^2$

$\sf 289x^2+578x+64y^2-512y-17183=0$

Veja que:

• $\sf (17x+17)^2=289x^2+578x+289$

• $\sf (8y-32)^2=64y^2-512y+1024$

Somando $\sf 17183+289+1024=18496$ a ambos os membros da equação:

$\sf 289x^2+578x+64y^2-512y-17183+18496=18496$

$\sf 289x^2+578x+64y^2-512y+1313=18496$

$\sf 289x^2+578x+64y^2-512y+289+1024=18496$

$\sf 289x^2+578x+289+64y^2-512y+1024=18496$

$\sf (17x+17)^2+(8y-32)^2=136^2$

$\sf [17\cdot(x+1)]^2+[8.(y-4)]^2=136^2$

$\sf 17^2\cdot(x + 1)^+8^2\cdot(y-4)^2=136^2$

Dividindo os dois lados por 136²:

$\sf \dfrac{17^2\cdot(x+1)^2}{136^2}+\dfrac{8^2\cdot(y-4)^2}{136^2}=\dfrac{136^2}{136^2}$

$\sf \dfrac{17^2\cdot(x+1)^2}{17^2\cdot8^2}+\dfrac{8^2\cdot(y-4)^2}{8^2\cdot17^2}=1$

$\sf \dfrac{(x+1)^2}{8^2}+\dfrac{(y-4)^2}{17^2}=1$

$\sf \dfrac{(y-4)^2}{17^2}+\dfrac{(x+1)^2}{8^2}=1$

$\sf \dfrac{(y-y_C)^2}{a^2}+\dfrac{(x-x_C)^2}{b^2}=1$

Logo:

• $\sf x_C=-1,~y_C=4~\Rightarrow~C(-1,4)~~(centro)$

• $\sf a=17~\Rightarrow~2a=34~~(eixo~maior)$

• $\sf b=8~\Rightarrow~2b=16~~(eixo~menor)$

$\sf b^2+c^2=a^2$

$\sf 8^2+c^2=17^2$

$\sf 64+c^2=289$

$\sf c^2=289-64$

$\sf c^2=225$

$\sf c=\sqrt{225}$

$\sf c=15$

A excentricidade dessa elipse é:

$\sf \dfrac{c}{a}=\dfrac{15}{17}$