What is the vertex and focus of the parabola described by #2x^2-5x+y+50=0#?

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Answer 1 · 2017-03-11T18:13:27

The vertex is #V=(5/4,-375/8)#
The focus is #F=(5/4,-376/8)#
The directrix is #y=-374/8#

Let's rewrite this equation and complete the squares

#2x^2-5x+y+50=0#

#2x^2-5x=-y-50#

#2(x^2-5/2x)=-(y+50)#

#(x^2-5/2x+25/16)=-1/2(y+50)#

#(x-5/4)^2=-1/2(y+50-25/8)#

#(x-5/4)^2=-1/2(y+425/8)#

We compare this equation to

#(x-a)^2=2p(y-b)#

The vertex is #V=(a,b)=(5/4,-375/8)#

#p=-1/4#

The focus is #F=(5/4,b+p/2)=(5/4,-376/8)#

The directrix is #y=b-p/2=-375/8+1/8=-374/8#

graph{(2x^2-5x+y+50)(y+374/8)((x-5/4)^2+(y+375/8)^2-0.001)=0 [-1.04, 7.734, -48.52, -44.13]}