diketahui hiperbola x2-16y2-4x-32y-28=0 tentukan koordinat fokus dan puncak hiperbola

[tex]x^2-16y^2-4x-32y-28=0 \\ x^2-4x+4-16x-32y-16-28=4-16 \\ (x^2-4x+4)-16(y^2+2y+1)=16 \\ (x-2)^2-16(y+1)^2=16 \\ \displaystyle \frac{(x-2)^2}{16}-\frac{(y+1)^2}{1}=1 \\ a^2=16 \\ a=4 \\ $Pusat : $(2,-1) \\ $Hiperbola horizontal$ \\ $Puncak : $(2-a,-1),(2+a,-1) \\ $Puncak : $(-2,-1),(6,-1) \\ c=\sqrt{a^2+b^2}=\sqrt{16+1}=\sqrt{17} \\ $Fokus : $(2-c,-1),(2+c-1) \\ $Fokus : $(2-\sqrt{17},-1),(2+\sqrt{17},-1)[/tex]