Find the Properties y^2+6y-2x+13=0

Problem

y2+6*y−2*x+13=0

Solution

1. Identify the type of conic section by observing that only one variable, y is squared, which indicates the equation represents a parabola opening horizontally.

2. Rearrange the equation to isolate the terms involving y on one side and move the x and constant terms to the other side.

y2+6*y=2*x−13

3. Complete the square for the y terms by adding (6/2)2=9 to both sides of the equation.

y2+6*y+9=2*x−13+9

4. Factor the left side into a perfect square and simplify the right side.

(y+3)2=2*x−4

5. Factor out the coefficient of x on the right side to put the equation into the standard form (y−k)2=4*p*(x−h)

(y+3)2=2*(x−2)

6. Determine the vertex (h,k) by identifying the values from the standard form.

(h,k)=(2,−3)

7. Calculate the value of p by setting 4*p equal to the coefficient of the linear term.

4*p=2

p=0.5

8. Find the focus using the formula (h+p,k) since the parabola opens to the right.

Focus=(2+0.5,−3)=(2.5,−3)

9. Find the directrix using the formula x=h−p

x=2−0.5

x=1.5

10. Identify the axis of symmetry which is the horizontal line passing through the vertex.

y=−3

Final Answer

Vertex: *(2,−3), Focus: *(2.5,−3), Directrix: *x=1.5, Axis of Symmetry: *y=−3

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