Question

determine a equação reduzida da hipérbole

Answer 1

$9y^2-16x^2-160x-544=0\\ \\ 9y^2-16(x^2+10+34)=0\\ \\ 9y^2-16[(x+5)^2+9]=0\\ \\ 9y^2-16(x+5)^2=16 \times 9\\ \\ \displaystyle\frac{y^2}{16}-\frac{(x+5)^2}{9}=1$

Centro (-5,0)
Hipérbola: vertical
a=3
b=4

Answer 2

$9 y^{2} - 16 x^{2} - 160x - 544=0--\ \textgreater \ 9 y^{2} - 16 (x^{2} +10x )=544 ;$ $9y^{2} - 16 (x^{2} +10x+25 )=544 - 400;9 y^{2} - 16 ( x + 5)^{2} = 144 :(144);$ $\frac{ y^{2} }{16} - \frac{(x + 5)^{2} }{9} =1;$