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]}

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