SOLUTION: trying to find the vertex, focus, latus rectum, directrix, axis of symmetry for the following equation -2(x+3)=(y-1)^2

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Question 1133782: trying to find the vertex, focus, latus rectum, directrix, axis of symmetry for the following equation
-2(x+3)=(y-1)^2
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
focus, latus rectum, directrix, axis of symmetry
-2%28x%2B3%29=%28y-1%29%5E2+
4p%28x-h%29=%28y-k%29%5E2 is the standard equation for a right-left facing parabola with vertex at (h,k) and a focal length abs%28p%29
Rewrite-2%28x%2B3%29=%28y-1%29%5E2+ in the standard form
as you can see 4p=-2=>p=-1%2F2
+4%28-1%2F2%29%28x-%28-3%29%29=%28y-1%29%5E2
Therefore parabola properties are:
(h,k) =(-3,1)
p=-1%2F2
Parabola is symmetric around the x-axis and so the focus lies a distance p from the center (-3%2Bp, 1 ) along the x-axis
(-3-1%2F2, 1 )....plug in p=-1%2F2
focus: (-7%2F2, 1 )
directrix is
x=-3-p
x+=+-3-%28-1%2F2%29
x+=+-6%2F2%2B1%2F2
x+=+-5%2F2

latus rectum:
-2%28x%2B3%29=%28y-1%29%5E2+............switch sides
%28y-1%29%5E2=-2%28x%2B3%29+......solve for y
y-1=+sqrt%28-2%28x%2B3%29+%29
y=+sqrt%28-2%28x%2B3%29+%29%2B1
y=+sqrt%282%28-x-3%29+%29%2B1
=> solutions:
y=+sqrt%282%29sqrt%28-x-3%29+%2B1
y=+1-sqrt%282%29sqrt%28-x-3%29+

Axis of symmetry is a line parallel to the x -axis which intersects the vertex:
y=1