Question 724278
How do i find the focus of the problem (y+3)^2=x-2
This is an equation of a parabola that opens rightward.
Its basic form: (y-k)^2=4p(x-h), (h,k)=(x,y) coordinates of the vertex
For given equation:
vertex: (2,-3)
axis of symmetry: y=3
4p=1
p=1/4
focus: (9/4,-3) (p-distance to the right of the vertex on the axis of symmetry)