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Question 967923: The equation of a parabola is given.
a. Write the equation of the parabola in standard form.
b. Identify the vertex, focus, and directrix.
Express numbers in exact simplest form.
16y^2-56y-16x+17=0
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! The equation of a parabola is given.
a. Write the equation of the parabola in standard form.
b. Identify the vertex, focus, and directrix.
Express numbers in exact simplest form.
16y^2-56y-16x+17=0
16(y^2-(56/16)y)-16x+17=0
16(y^2-(7/2)y+49/16)-49-16x+17=0
16(y-7/4)^2-16x-32=0
16(y-7/4)^2-16(x+2)=0
16(y-7/4)^2=16(x+2)
divide both sides by 16
(y-7/4)^2=(x+2)
this is an equation of a parabola that opens rightward
Its basic form: (y-k)^2=4p(x-h), (h,k)=coordinates of the vertex
vertex: (-2, 7/4)
axis of symmetry: y=7/4
p=1
focus: (-1, 7/4) (p-distance to the right of vertex on the axis of symmetry)
directrix: x=-3 (p-distance to the left of vertex on the axis of symmetry)
vertex form of equation for given parabola: x=(y-7/4)^2-2
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