How do you identify the vertex, focus, directrix and the length of the latus rectum and graph #x=y^2-14y+25#?

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Answer 1 · 2017-03-05T10:49:46

The vertex is #V=(-24,7)#
The focus is #F=(-95/4,7)#
The directrix is #x==-97/4#

Let's rewrite this equation and complete the squares

#x=y^2-14y+25#

#x-25=y^2-14y#

#(x-25+49)=y^2-14y+49#

#x+24=(y-7)^2#

#(y-7)^2=x+24#

We compare this equation to

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

The vertex is #V=(a,b)=(-24,7)#

#p=1/2#

The focus is #F=(a+p/2,b)=(-95/4,7)#

The directrix is #x=a-p/2=-24-1/4=-97/4#

graph{(x-y^2+14y-25)(y-1000(x+97/4))=0 [-27.216, -13.17, 3.35, 10.37]}