write this equation of parabola in its canonical form:x^2-6x-4y+13=0

Question

And hence determine its foci vertices  directrice.

Answer 1

If by canonical form you mean y-k=(1/4a)(x-h) where (h,k) is the vertex and a is the distance between the vertex and focus and directrix line, then we can rewrite the equation as:

(x²-6x+9)-9-4y+13=0, (x-3)²+4-4y=0, 4(y-1)=(x-3)², y-1=¼(x-3)², so the vertex is (3,1) and a=1.

The focus has the same x coord as the vertex: (h,k+a)=(3,2) and the directrix line is y=k-a=1-1=0 the x-axis, y=0.