Question

Find the center, foci, and vertices of the hyperbola. 16y^2 − x^2 + 2x + 128y + 175 = 0

Answer 1

\[16y ^{2}-x ^{2}+2x+128y+175=0\]\[16y ^{2}+128y-x ^{2}+2x=-175\]\[16(y ^{2}+8y)-x ^{2}+2x=-175\]Complete the square on the y and x variables.
\[16[(y+4)^{2}-16]-1[(x-1)^{2}-1]=-175\]\[16(y+4)^{2}-(x-1)^{2}=80\]\[\frac{ 16 }{ 80 }(y+2)^{2}-\frac{ (x-1)^{2} }{ 80 }=1\]\[\frac{ (y+4)^{2} }{ 5 }-\frac{ (x-1)^{2} }{ 80}=1\]Center is (4,1) I'll let you find the vertices and foci