Question

How do you find the vertex, focus and directrix of `x^2-10x-8y+33=0`?

Answer 1

Vertex is at `(5,1)` , focus is at `(5,3)` and
directrix is `-1` .

Explanation:

`x^2-10x-8y+33=0 or 8y = x^2-10x+33` or

`8y = (x^2-10x+25)-25+33 or 8y = (x-5)^2 +8` or

`y= 1/8(x-5)^2 +1`. Comparing with standard equation `y= a(x-h)^2+k ; (h,k)`being vertex , we get here `h=5 ,k =1 , a= 1/8`. So vertex is at `(5,1)`. The parabola opens upwards since `a>0`

Vertex is at equidistant from focus and directrix. The distance of vertex from directrix is `d = 1/(4|a|) =1/(4*1/8)=2` and directrix is below the vertex.

So directrix is `y=(1-2)= -1` . Focus is above the vertex and at `5,(1+2) or (5,3)`
graph{x^2-10x-8y+33=0 [-20, 20, -10, 10]} [Ans]