Find the Properties y^2+6y+8x+25=0

Problem

y2+6*y+8*x+25=0

Solution

1. Identify the conic section by observing that only one variable, y is squared. This indicates the equation represents a parabola that opens horizontally.

2. Complete the square for the y terms to rewrite the equation in standard form. Move the terms not containing y to the right side of the equation.

y2+6*y=−8*x−25

3. Add the square of half the coefficient of y to both sides. Half of 6 is 3 and 3=9

y2+6*y+9=−8*x−25+9

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

(y+3)2=−8*x−16

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

(y+3)2=−8*(x+2)

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

h=−2

k=−3

Vertex=(−2,−3)

7. Calculate the value of p by setting 4*p equal to the coefficient in front of the x term.

4*p=−8

p=−2

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

Focus=(−2+(−2),−3)=(−4,−3)

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

x=−2−(−2)

x=0

10. Identify the axis of symmetry using the formula y=k

y=−3

Final Answer

(y+3)2=−8*(x+2)⇒Vertex: *(−2,−3), Focus: *(−4,−3), Directrix: *x=0, Axis of Symmetry: *y=−3

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