Answer:
Vertex is (3,3)
Focus (3,6)
Directrix y=0
Step-by-step explanation:
We are given that an equation of parabola
[tex]12y=x^2-6x+45[/tex]
Subtract 45 on bot sides then we get
[tex]12y-45=x^2-6x[/tex]
To make complete square on right side
We add 9 on both sides
[tex]12y-45+9=x^2-6y+9[/tex]
[tex]12y-36=(x-3)^2[/tex] ([tex](a-b)^2=a^2+b^2-2ab,(x-3)^2=x^2-6x+9)[/tex]
[tex]12(y-3)=(x-3)^2[/tex]
Compare it with the equation of parabola along y- axis
[tex](x-h)^2=4p(y-k)[/tex]
Then, we get h=3,k=[tex]3[/tex]
[tex]4p=12[/tex]
[tex]p=\frac{12}{4}=3[/tex]
Vertex of parabola =(h,k)=(3,3)
Focus of parabola=(x,p+k)=(3,3+3)=(3,6)
Equation of directrix of parabola ,y=k-p=3-3=0
