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

1 Answer
Binayaka C.
Jun 17, 2017

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]

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