Find the Properties x^2+8x=4y-8

Problem

x2+8*x=4*y−8

Solution

1. Identify the type of conic section by observing that only one variable, x is squared. This indicates the equation represents a parabola that opens vertically.

2. Complete the square for the x terms by taking half of the coefficient of x squaring it, and adding it to both sides.

8/2=4

4=16

x2+8*x+16=4*y−8+16

3. Simplify both sides of the equation to put it into the standard form (x−h)2=4*p*(y−k)

(x+4)2=4*y+8

(x+4)2=4*(y+2)

4. Determine the vertex (h,k) from the standard form.

h=−4

k=−2

Vertex=(−4,−2)

5. Calculate the focal length p by setting the coefficient of the linear term equal to 4*p

4*p=4

p=1

6. Find the focus by adding p to the ycoordinate of the vertex, since the parabola opens upward (p>0.

Focus=(h,k+p)

Focus=(−4,−2+1)=(−4,−1)

7. Find the directrix by subtracting p from the ycoordinate of the vertex.

y=k−p

y=−2−1=−3

8. Identify the axis of symmetry which passes through the vertex and focus.

x=h

x=−4

Final Answer

Vertex: *(−4,−2), Focus: *(−4,−1), Directrix: *y=−3, Axis of Symmetry: *x=−4

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