SOLUTION: what is the latus rectum of this equation? y^2-12y+16x+36=0 standard form = (y-6)^2=-16x

SOLUTION: what is the latus rectum of this equation? y^2-12y+16x+36=0 standard form = (y-6)^2=-16x

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Question 817626: what is the latus rectum of this equation?
y^2-12y+16x+36=0
standard form = (y-6)^2=-16x
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
what is the latus rectum of this equation?
y^2-12y+16x+36=0
standard form = (y-6)^2=-16x
This is an equation of a parabola that opens left
Its basic form of equation: (y-k)^2=4p(x-h)
latus rectum, aka, focal width=4p=16
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