SOLUTION: Complete the square, then graph and identify the vertex, focus, directrix, and endpoints of the latus rectum for the equation -14x+2y^2-8y=20

Question 615731: Complete the square, then graph and identify the vertex, focus, directrix, and endpoints of the latus rectum for the equation -14x+2y^2-8y=20
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Complete the square, then graph and identify the vertex, focus, directrix, and endpoints of the latus rectum for the equation -14x+2y^2-8y=20
-14x+2y^2-8y=20
complete the square
2(y^2-4y+4)=20+14x+8
2(y-2)^2=14x+28
divide by 2
(y-2)^2=7x+14
(y-2)^2=7(x+2)
This is an equation of a parabola that opens rightwards.
Its standard form: (y-k)^2=4px, (h,k)=(x,y) coordinates of the vertex
For given equation:(y-2)^2=7(x+2)
vertex:(-2,2)
axis of symmetry: y=2
4p=7
p=7/4
Focus: (-2+p,2)=(-2+7/4,2)=(-1/4,2) (p distance to the right of the vertex on the axis of symmetry)
Directrix: x=(-2-p)=(-2-7/4)=-15/4 (p distance to the left of the vertex on the axis of symmetry)
latus rectum:
length of latus rectum=4p=7
2p=7/2
end points: (-1/4,2�2p)
=(-1/4,2�7/2)
=(-1/4,-1.5) and (-1/4,5.5)
see graph below:
y=(7(x+2))^.5+2

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