SOLUTION: Find the latus rectum of given equation y^2+8x-6y+25=0

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Question 1115791: Find the latus rectum of given equation y^2+8x-6y+25=0
Found 2 solutions by josgarithmetic, stanbon:
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
y%5E2-6y=-8x-25
y%5E2-6y%2B9=-8x-25%2B9
%28y-3%29%5E2=-8x-16
%28y-3%29%5E2=-8%28x%2B2%29

vertex (-2,3)

Distance between vertex and focus p;
-8=-4p
p=2
Focus is two unit to the left of (-2,3)
Focus is at (-4,3).

LATUS RECTUM?
y=3%2B-+sqrt%28-8%28x%2B2%29%29
x coordinate of focus is -4.
y=3%2B-+sqrt%28-8%28-4%2B2%29%29
y=3%2B-+4
system%28y=-1%2Cor%2C7%29-------distance across the parabola at x=-4, is 8.
Latus Recturm is highlight%288%29.

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Find the latus rectum of given equation y^2+8x-6y+25=0
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y^2-6y+9 = -8x-16
-8(x+2) = (y-3)^2
----
4p = -8
p = -2
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You have a parabola opening to the left with vertex at (-2,3)
The focus is 2 to the left of (-2,3) and corresponds to the
line y = -4.
The latus rectum is a vertical line segment passing thru the focus.
Length of the latus rectum = |4p| = 8
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Cheers,
Stan H.
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