y^2 = x-1 (x-3)^2 + y^2 = 4

Langsung substitusi aja,
[tex]y^2=x-1 \\ (x-3)^2+y^2=4 \\ $Menjadi,$ \\ (x-3)^2+(x-1)=4 \\ x^2-6x+9+x-1=4 \\ x^2-5x+4=0 \\ (x-1)(x-4)=0 \\ x=1$ dan $x=4 \\ $Untuk $x=1, \\ y^2=x-1=1-1=0 \\ y=0 \\ $Untuk $x=4 \\ y^2=x-1=4-1=3 \\ y=\pm\sqrt{3} \\ \\ $HP$=(1,0),(4,\sqrt{3}),(4,-\sqrt{3})[/tex]