Find the Properties -2y^2+x-4y+1=0

Problem

−2*y^2+x−4*y+1=0

Solution

1. Identify the type of conic section. Since the equation contains a y^2 term but no x^2 term, it represents a parabola that opens horizontally.

2. Isolate the x terms on one side to prepare for completing the square for y

x=2*y^2+4*y−1

3. Factor out the coefficient of y^2 from the y terms.

x=2*(y^2+2*y)−1

4. Complete the square for the expression inside the parentheses. Add and subtract (2/2)^2=1 inside the parentheses.

x=2*(y^2+2*y+1−1)−1

x=2*(y+1)^2−2−1

x=2*(y+1)^2−3

5. Rewrite in the standard form (y−k)^2=4*p*(x−h) or x−h=a*(y−k)^2

x+3=2*(y+1)^2

(y+1)^2=1/2*(x+3)

6. Determine the vertex (h,k) from the standard form (y−k)^2=4*p*(x−h)

h=−3

k=−1

Vertex=(−3,−1)

7. Calculate the value of p using 4*p=1/2

p=1/8

8. Find the focus (h+p,k)

Focus=(−3+1/8,−1)

Focus=(−23/8,−1)

9. Find the directrix x=h−p

x=−3−1/8

x=−25/8

10. Identify the axis of symmetry y=k

y=−1

Final Answer

Vertex: *(−3,−1), Focus: *(−23/8,−1), Directrix: *x=−25/8, Axis of Symmetry: *y=−1

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