Problem 36 - College Algebra with Corequisite Support

Question

For the following exercises, graph the parabola, labeling the focus and the directrix. \(-5(x+5)^{2}=4(y+5)\)

Step-by-Step Solution

Answer

Vertex: (-5, -5); Focus: (-5, -\frac{85}{16}); Directrix: y = -\frac{75}{16}.

Step 1: Rewrite the Equation in Standard Form

The given equation is \(-5(x+5)^2 = 4(y+5)\). First, divide both sides by 4 to isolate \((y+5)\): \[-\frac{5}{4}(x+5)^2 = (y+5)\]. Rewrite it as: \[(y + 5) = -\frac{5}{4}(x + 5)^2\]. This is the standard form of a parabola opening downwards.

Step 2: Identify the Vertex

The vertex form of the equation is \((x-h)^2 = -\frac{4p}{a} (y-k)\) for a parabola opening up or down. Comparing with \((y + 5) = -\frac{5}{4}(x + 5)^2\), we find that the vertex is at \(h = -5\) and \(k = -5\), giving us the point \((-5, -5)\).

Step 3: Determine the Direction and Distance to the Focus

The equation \((y + 5) = -\frac{5}{4}(x + 5)^2\) implies the parabola opens downwards because of the negative coefficient. For the focus, calculate \(p\) by setting \(-\frac{5}{4} = -4p\), thus \(p = \frac{5}{16}\). The focus is located \(p\) units below the vertex, so the focus is at \((-5, -5 - \frac{5}{16}) = (-5, -\frac{85}{16})\).

Step 4: Find the Equation of the Directrix

The directrix is \(p\) units above the vertex at \(y = k + p\). Substituting the values, we get \(y = -5 + \frac{5}{16} = -\frac{75}{16}\). Thus, the equation of the directrix is \(y = -\frac{75}{16}\).

Step 5: Graph the Parabola

Plot the vertex at \((-5, -5)\). Mark the focus at \((-5, -\frac{85}{16})\) and draw the directrix line at \(y = -\frac{75}{16}\). Sketch the parabola opening downwards from the vertex. Make sure the vertex is midway between the focus and directrix.

Key Concepts

Focus of a parabola

The focus of a parabola is a special point used to define the curve and is one of the basic ways a parabola is geometrically defined. The focus lies on the symmetry axis of the parabola. When considering upward or downward-opening parabolas, it is either above or below the vertex, respectively.

For the given parabola, \[(y + 5) = -\frac{5}{4}(x + 5)^2,\]which opens downward, you can find the focus by calculating the distance \(p\) from the vertex, using the formula \(-\frac{5}{4} = -4p\), giving \(p = \frac{5}{16}\).

• The vertex is at \((-5, -5)\).
• Focus is \(p\) units from the vertex, so it’s \((-5, -5 - \frac{5}{16})\).
• The focus is thus located at \((-5, -\frac{85}{16})\).

The focus makes sure that the parabola’s points are equidistant to it and the directrix. This relationship is crucial for graphing and understanding the geometry of a parabola.

Directrix

The directrix of a parabola is a fixed line used in the geometric definition of the parabola. It is equally important as the focus, working together to define the parabola. For a downward-opening parabola, the directrix is a horizontal line positioned above the vertex.

In our example, the given equation is \[-\frac{5}{4}(x+5)^2 = (y+5)\]. We determined that \(p = \frac{5}{16}\). Therefore, the directrix is a line that is \(p\) units above the vertex at \(y = k + p\), where \(k\) is the y-coordinate of the vertex.

• Vertex: \((-5, -5)\)
• Directrix: \(y = -5 + \frac{5}{16} = -\frac{75}{16}\)

The line \(y = -\frac{75}{16}\) acts as the directrix and completes the geometric definition alongside the focus. Every point on the parabola is equidistant from the focus and this directrix. This is how you verify the correctness of your derived equation and plot.

Vertex (of a parabola)

The vertex of a parabola is the point where the curve changes direction. It is the maximum or minimum point for the parabola, depending on whether it opens upwards or downwards.

To determine the vertex, we compare the equation to the standard form \[(x-h)^2 = a(y-k)\] for parabolas opening up or down. Our equation is \[-\frac{5}{4}(x+5)^2 = (y+5)\]. By comparing, we deduce that the vertex \((h, k)\) is at:

• \(h = -5\)
• \(k = -5\)

Thus, the vertex is located at \((-5, -5)\).

The vertex is pivotal for graphing because it is the symmetrical midpoint between the focus and the directrix. This helps establish the parabola's orientation and exact shape. Understanding the vertex aids in graphing and gathering insights into the movement and transformation of parabolas.

Graphing

Graphing a parabola involves plotting the vertex, focus, and directrix to visualize the curve's path. Graphically, the vertex is your starting point, while the focus and directrix help in drawing the curve accurately.

In our exercise, after determining the vertex at \((-5, -5)\), focus at \((-5, -\frac{85}{16})\), and directrix at \(y = -\frac{75}{16}\), you can start drawing the parabola opening downward. Here’s the step-by-step process:

• Start by plotting the vertex \((-5, -5)\).
• Mark the focus point below the vertex for downward-opening parabola.
• Draw the horizontal directrix line above.
• Sketch the parabola symmetrically ensuring every point is equidistant from the focus and the directrix.

Graphing allows you to see the relationship between algebraic equations and geometric figures. Visualizing parabolas helps solidify your understanding and creates a lasting impression of their defining properties.