Question

Graph the parabola. Find the vertext, focus, directix and the endpoints of the latus.

4y^2+2x=0

Answer 1

Given equation 4y^2+2x = 0

4y^2 = -2x

y^2 = -2x/4

y^2 = -1/2x

y^2 = 4(-1/8)x

Compare the standard equation of parabola y^2 = 4ax

Vettrtex = (0,0)

Focus = (a,0) = (-1/8,0)

Equation of drectrix x = -a

x = -(-1/8)

x = 1/8

the length of parabola locus rectum is 4p.where p is distance from the focus to tht vetrex.

p = √(0-1/8)^2+(0-0)^2 = √1/64 = 1/8 

end points of latus  is (-1/8,1/4)(-1/8,-1/4)

 

 

Answer 2

Note : The standard form of parabola equation is (y - k)^2 = 4p (x - h), where (h, k) = vertex and p = directed distance from vertex to focus.

The equation is 4y² + 2x = 0.

Rewrite the equation as y² = - x/2.

(y - 0)² = (- 1/2)(x - 0).

The above equation represent the parabola.

Compare the above equation with the standard form of parabola equation is (y - k)^2 = 4p (x - h), where (h, k) = vertex and p = directed distance from vertex to focus.

• Vertex (h, k) = (0, 0)

The length of parabola locus rectum is 4p.where p is distance from the focus to tht vetrex.

p = √(0-1/8)^2+(0-0)^2 = √1/64 = 1/8

• p = directed distance from vertex to focus = 1/8.

Focus is p distance to the left of vertex on the axis of symmetry.

Focus = (h + p, k)

• Focus = (0 - 1/8, 0) = (- 1/8, 0).
• End points of latus  is (-1/8, 1/4)(-1/8, -1/4).

Make the table of values to find ordered pairs that satisfy the equation.

Choose values for y and find the corresponding values for x.

y	x = - 2y2	(x, y)
- 1	x = - 2(- 1)2 = - 2	(- 2, - 1)
- 0.5	x = - 2(- 0.5)2 = - 0.5	(- 0.5, - 0.5)
0	x = - 2(0)2 = 0	(0, 0)
0.5	x = - 2(0.5)2 = - 0.5	(- 0.5, 0.5)
1	x = - 2(1)2 = - 2	(- 2, 1)

Graph :

1.Draw a coordinate plane.

2.Plot the coordinate points.

3.Then sketch the graph, connecting the points with a smooth curve.

Comments

The standard form of parabola 4y^2+2x=0 is (y - 0)^2 = 4(- 1/8) (x - 0).

h = 0, k = 0 and p = - 1/8.

Directrix is x = h - p = 0 - (- 1/8) = 1/8.