plis, minta bantuannya, makasih sebelumnya. persamaan garis singgung hiperbola 8y^2-4x^2-16y-24x-60=0 di titik (5,-5) adalah...

[tex]8y^2-4x^2-16y-24x-60=0 \\ 8y^2-16y+8-(4x^2+24x+36)-32=0 \\ 8(y^2-2y+1)-4(x^2+6x+9)=32 \\ 8(y-1)^2-4(x+3)^2=32 \\ 2(y-1)^2-(x+3)^2=8 \\ \displaystyle \frac{(y-1)^2}{4}-\frac{(x+3)^2}{8}=1 \\ \\ $Garis singgung:$ \\ \displaystyle \frac{(y-y_1)(y'-y_1)}{a^2}-\frac{(x-x_1)(x'-x_1)}{b^2}=1 \\ \\ \frac{(y-1)(-5-1)}{4}-\frac{(x+3)(5+3)}{8}=1 \\ \frac{-6(y-1)}{4}-\frac{8(x+3)}{8}=1 \\ -\frac{3}{2}(y-1)-(x+3)=1 \\ -3(y-1)-2(x+3)=2 \\ -3y+3-2x-6=2 \\ 2x+3y+5=0[/tex]