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