Problem 41 - College Algebra with Corequisite Support

Question

For the following exercises, graph the parabola, labeling the focus and the directrix. \(y^{2}-8 x+10 y+9=0\)

Step-by-Step Solution

Answer

Vertex: (-2, -5), Focus: (0, -5), Directrix: x = -4.

Step 1: Identifying the Standard Form

The given equation of the parabola is \(y^2 - 8x + 10y + 9 = 0\). To graph the parabola, we first need to rewrite it in a standard form \((y-k)^2 = 4p(x-h)\). This form easily shows us the vertex, focus, and directrix of the parabola.

Step 2: Completing the Square

Starting with the equation \(y^2 + 10y = 8x - 9\), we complete the square on the \(y\)-terms. We do that by taking half of 10, which is 5, squaring it to get 25, and adding to both sides:\[y^2 + 10y + 25 = 8x - 9 + 25\]This simplifies to:\((y+5)^2 = 8x + 16\).

Step 3: Expressing in Vertex Form

We now want to isolate \(x\):\((y+5)^2 = 8(x+2)\).This equation is now in the form \((y-k)^2 = 4p(x-h)\), where \(h = -2\), \(k = -5\), and \(4p = 8\). Thus, \(p = 2\).

Step 4: Identifying the Vertex

From the vertex form \((y+5)^2 = 8(x+2)\), we identify the vertex of the parabola as \((-2, -5)\).

Step 5: Finding the Focus

The value of \(p\) is 2. For a horizontal parabola (which opens in the x-direction), the focus is \(p\) units to the right if \(p\) is positive (and left if negative) from the vertex. The focus is at \((-2 + 2, -5) = (0, -5)\).

Step 6: Finding the Directrix

The directrix of the parabola, being a vertical line for a horizontally opening parabola, is \(x = h - p = -2 - 2 = -4\).

Step 7: Graphing the Parabola

Plot the vertex at \((-2, -5)\), the focus at \((0, -5)\), and the directrix as the vertical line \(x = -4\). The parabola, opening towards the right, will be symmetric around the line \(y = -5\).

Key Concepts

Graphing Parabolas

When graphing a parabola, understanding its form is crucial. Parabolas can open up, down, left, or right, and their shapes are determined by their equations. The standard form for a parabola that opens to the side is

• \((y - k)^2 = 4p(x - h)\)

The standard form makes it easier to identify the important features of the parabola such as its vertex, focus, and directrix.

To graph a parabola, you need to:

• Identify the vertex, which sets the center of the parabola.
• Plot the focus, which helps in determining the parabola's "width" and how it opens.
• Draw the directrix, aiding in understanding the symmetry and direction of the parabola.

Using these elements, you can sketch how the parabola sits on the coordinate plane.

Vertex of a Parabola

The vertex of a parabola is a critical point that defines its shape and position. You can think of it as the parabola's 'turning point' or the center from which it opens outward.

For parabolas in the standard side-opening form,

• \((y-k)^2 = 4p(x-h)\)

The vertex is located at the coordinates

• \((h, k)\)

This key position determines the axis of symmetry for the parabola.

In the given equation, through completing the square, the vertex is found to be at

• \((-2, -5)\)

This point tells us that this particular parabola is centered here and opens sideways, depending on the direction specified by the value of \(p\).

Focus and Directrix of a Parabola

The focus and directrix are fundamental aspects of a parabola that define its 'width' and direction.

• The focus is a point inside the parabola where all the lateral rays reflect to when you visualize the parabola as a physical, reflective curve.
• For our equation, the focus can be identified as
  \((h+p, k) = (0, -5)\)

The directrix is a line outside the parabola that serves as a 'mirror' for the focus. It aids in maintaining the parabolic curve's symmetry.

• Our directrix here is the vertical line
  \(x = -4\)

Having both these elements allows us to perfectly understand how wide and in which direction the parabola opens.

Completing the Square

Completing the square is a method used to simplify quadratic equations, transforming them into a form that reveals the parabola's key features.

To complete the square, focus on one variable and rearrange the equation. For instance, if we start with

• \(y^2 + 10y = 8x - 9\)

we extract the constant term that results in a perfect square trinomial. Taking half the coefficient of \(y\), squaring it, we add that to both sides to maintain equality:

• \((y+5)^2 = 8x + 16\)

This process effectively rearranges the equation to highlight the vertex, after which you can identify the position

• \((x+2)\)

This completed square provides a more intuitive, simpler form that helps us easily pinpoint the parabola's vertex and directions for graphing.