Math Problem Statement

X2+2x+4y-11=0 equation of a parabola in general form, make it in standard form, opening of the graph, vertex, length of lactus rectum, value of p, focus, endpoints of LR, directrix, axis of symmetry

Solution

The given equation is:

x2+2x+4y11=0x^2 + 2x + 4y - 11 = 0

Let's transform this into the standard form for a parabola, and then find the required properties like the opening direction, vertex, length of the latus rectum, value of pp, focus, endpoints of the latus rectum, directrix, and axis of symmetry.

Step 1: Rearranging the equation

We can rewrite the equation by grouping the xx-terms on the left and isolating the yy-terms on the right:

x2+2x=114yx^2 + 2x = 11 - 4y

Now, we need to complete the square for the xx-terms.

Step 2: Completing the square

To complete the square for x2+2xx^2 + 2x, take half of the coefficient of xx (which is 2), square it, and add it to both sides:

x2+2x+1=114y+1x^2 + 2x + 1 = 11 - 4y + 1

This simplifies to:

(x+1)2=124y(x + 1)^2 = 12 - 4y

Now isolate yy:

(x+1)2=4(y3)(x + 1)^2 = -4(y - 3)

Step 3: Standard form of the parabola

The equation is now in the standard form of a parabola:

(x+1)2=4(y3)(x + 1)^2 = -4(y - 3)

This is a vertical parabola, and it opens downward because of the negative sign in front of the 4(y3)4(y - 3).

Step 4: Vertex

The vertex of the parabola is at the point (1,3)(-1, 3), derived from the standard form (xh)2=4p(yk)(x - h)^2 = 4p(y - k), where (h,k)(h, k) is the vertex.

Step 5: Value of pp

From the equation (x+1)2=4(y3)(x + 1)^2 = -4(y - 3), we have 4p=44p = -4, so p=1p = -1.

Step 6: Focus

The focus is located at (h,k+p)(h, k + p). Since h=1h = -1, k=3k = 3, and p=1p = -1, the focus is at:

Focus=(1,31)=(1,2)\text{Focus} = (-1, 3 - 1) = (-1, 2)

Step 7: Directrix

The directrix is the line y=kpy = k - p. Therefore, the directrix is:

Directrix=y=3+1=4\text{Directrix} = y = 3 + 1 = 4

Step 8: Axis of symmetry

The axis of symmetry is the vertical line passing through the vertex, which is:

Axis of symmetry=x=1\text{Axis of symmetry} = x = -1

Step 9: Length of the latus rectum

The length of the latus rectum is given by 4p|4p|. Since p=1p = -1, the length of the latus rectum is:

Length of latus rectum=4(1)=4\text{Length of latus rectum} = |4(-1)| = 4

Step 10: Endpoints of the latus rectum

The endpoints of the latus rectum are located at (h±2p,k+p)(h \pm 2p, k + p). Using h=1h = -1, k=3k = 3, and p=1p = -1:

Endpoints of latus rectum=(1±2,2)=(3,2) and (1,2)\text{Endpoints of latus rectum} = (-1 \pm 2, 2) = (-3, 2) \text{ and } (1, 2)

Summary:

  • Opening: Downward
  • Vertex: (1,3)(-1, 3)
  • Focus: (1,2)(-1, 2)
  • Directrix: y=4y = 4
  • Axis of symmetry: x=1x = -1
  • Length of latus rectum: 4
  • Endpoints of latus rectum: (3,2)(-3, 2) and (1,2)(1, 2)

Would you like more details on any of these concepts?

Here are five related questions:

  1. How do you complete the square for other conic sections, like ellipses or hyperbolas?
  2. What is the significance of the value pp in parabolas?
  3. How do the properties of horizontal parabolas differ from vertical parabolas?
  4. Can you find the equation of the parabola given its focus and directrix?
  5. What are the differences between the general form and standard form of conic sections?

Tip: For parabolas, the sign of pp determines the direction of opening—positive for upward/rightward and negative for downward/leftward.

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Math Problem Analysis

Mathematical Concepts

Conic Sections
Parabolas
Coordinate Geometry

Formulas

Standard form of a vertical parabola: (x - h)^2 = 4p(y - k)
Length of the latus rectum: |4p|
Focus of the parabola: (h, k + p)
Directrix of the parabola: y = k - p
Axis of symmetry: x = h

Theorems

Parabola properties
Vertex form of a quadratic equation

Suitable Grade Level

Grades 9-12