SOLUTION: find the coordinates of the vertex, focus, and endpoints of the latus rectum. also find the equation of the directrix of 4x^2-16x-15y+1=0.

Question 949635: find the coordinates of the vertex, focus, and endpoints of the latus rectum. also find the equation of the directrix of 4x^2-16x-15y+1=0.
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
find the coordinates of the vertex, focus, and endpoints of the latus rectum. also find the equation of the directrix of 4x^2-16x-15y+1=0.
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basic form of equation for parabola that opens upward: (x-h)^2=4p(y-k), (h,k)=coordinates of the vertex.
complete the square:
4(x^2-4x+4)-16-15y+1=0
4(x-2)^2=15y+15
4(x-2)^2=15(y+1)
(x-2)^2=(15/4)(y+1)
vertex: (2,-1)
axis of symmetry: x=2
4p=15/4
p=15/16
focus: (2,-1/16) (p-distance above vertex on the axis of symmetry)
directrix :y=31/16 (p-distance below vertex on the axis of symmetry)
endpoints of latus rectum:
(x-2)^2=(15/4)(y+1)
y=-1/16
(x-2)^2=(15/4)(-1/16+1)
(x-2)^2=(15/4)(15/16)
(x-2)^2=(15^2/64)
take sqrt of both sides
x-2=�15/8
x=2�15/8=16/8�15/8=1/8, 31/8
endpoints of latus rectum: (1/8,-1/16) and (31/8, -1/16)

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