Problem 48 - Precalculus with Limits

Question

In Exercises 47-50, find the vertex, focus, and directrix of the parabola. Use a graphing utility to graph the parabola. \(x^2 - 2x + 8y + 9 = 0\)

Step-by-Step Solution

Answer

For the given parabola equation, the vertex is at the point (1, -1), the focus is at (1, 1) and the directrix is the line given by \(y = -3\).

Step 1: Rewrite the equation into standard form

The given parabolic equation is \(x^2 - 2x + 8y + 9 = 0\). By completing the square, we can write it in standard form which is \((x-h)^2 = 4a(y-k)\). On rearranging, we get \((x - 1)^2 = 8(y + 1)\) which is the required standard form.

Step 2: Find the vertex

From Step 1, the standard form of a parabolic equation tells us that (h, k) give us the vertex of the parabola. Hence, in this case, the vertex would be at the point (1, -1).

Step 3: Find the value of 'a'

The coefficient of (y-k) in the standard form, after divided by 4, gives the value of \'a\'. In this case, it is \(8/4 = 2\). This 'a' will be useful for determining the focus and directrix.

Step 4: Determine the focus

The focus of a parabola in this form is always a units above the vertex when the parabola opens upwards. Thus, with 'a' being 2, the focus would be at (1, -1+2) = (1, 1).

Step 5: Determine the directrix

The directrix will be the horizontal line crossing the y-axis at 'k-a'. Thus, in this case, the equation for the directrix would be \(y = -1 - 2 = -3\).

Key Concepts

Vertex

The vertex of a parabola is its highest or lowest point, depending on its orientation. It's like the peak if the parabola opens downward, or the valley if it opens upward. For the equation \((x-1)^2 = 8(y+1)\), the vertex is found by identifying \((h, k)\) in the standard form \((x-h)^2 = 4a(y-k)\). Here, \(h = 1\) and \(k = -1\), so the vertex is at \((1, -1)\). This point is crucial because it acts as the "turning point" of the parabola.

• The x-coordinate (\(h\)) shifts the parabola left or right.
• The y-coordinate (\(k\)) moves it up or down.

Understanding the vertex helps in sketching and analyzing the parabola's shape.

Focus

The focus of a parabola is a point from which distances to the curve reflect equally. It is one of the points that helps in defining the parabola's shape. In our equation \((x-1)^2 = 8(y+1)\), the focus is a distance \(a\) away from the vertex along the axis of symmetry. We've determined \(a = 2\) from the standard form. So, the focus is \(a\) units above \((h, k)\), resulting in \((1, 1)\).

• This point lies within the curve.
• It's always inside the parabola, along the "direction" it opens (up, down, left, right).

The focus's location helps to demonstrate the reflective property of parabolas, which is central to their practical applications.

Directrix

The directrix of a parabola is a line that, along with the focus, helps define the parabola's shape. It is perpendicular to the axis of symmetry. For our equation, the directrix is the line \(y = -3\), calculated as \(y = k - a\) where \(k = -1\) and \(a = 2\).

• It’s a fixed, straight line.
• Each point on the parabola is equidistant from the focus and the directrix.

Understanding the directrix helps visualize how parabolas reflect light and sound waves, which is essential in various technologies like satellite dishes and car headlights.

Standard Form of a Parabola

The standard form of a parabola is a key algebraic expression used to easily identify its specific properties. For vertical parabolas, it is \((x-h)^2 = 4a(y-k)\), and for horizontal ones, it’s \((y-k)^2 = 4a(x-h)\). In our exercise, the equation \((x-1)^2 = 8(y+1)\) represents a vertically oriented parabola.

• \((h, k)\) gives the vertex location.
• \(4a\) determines the distance between the vertex and the focus or directrix.
• If \(a\) is positive, the parabola opens upwards; if negative, it opens downwards.

Rewriting equations into this form simplifies the process of identifying key features like the vertex, focus, and directrix, making it easier to graph and comprehend the parabola's behavior.