Problem 54 - College Algebra with Modeling and Visualization

Question

Graph the parabola. Label the vertex, focus, and directrix. $$ \frac{1}{16}(x+4)^{2}=-(y-4) $$

Step-by-Step Solution

Answer

Vertex: (-4, 4); Focus: (-4, 0); Directrix: y = 8.

Step 1: Rearrange the Equation

Start by rearranging the given equation \[ \frac{1}{16}(x+4)^2 = -(y-4) \] into the general form of a vertical parabola, \[ (x-h)^2 = -4p(y-k), \]which in this case implies multiplying through by -1, so it becomes \[ (x+4)^2 = -16(y-4). \]

Step 2: Identify the Vertex

The vertex form of a parabola is usually specified as \[ (x-h)^2 = 4p(y-k) \]for vertical parabolas. Thus, compared with our equation \[ (x+4)^2 = -16(y-4), \]we see that \[ h = -4, \, k = 4. \] Therefore, the vertex is at \[ (-4, 4). \]

Step 3: Determine the Value of p

In the equation \[ (x+4)^2 = -16(y-4), \]the term \[ -16 \] is equivalent to \[-4p. \] Thus, we set \[-4p = -16 \] and solve for \(p\): \[ p = 4. \]

Step 4: Locate the Focus

For a parabola with a vertex at \[ (h, k) = (-4, 4) \]and a focus located \(p\) units from the vertex along the axis of symmetry, since \(p = 4\) and the parabola opens downwards (negative coefficient), the focus is located \[ (h, k-p) = (-4, 4-4) = (-4, 0). \]

Step 5: Find the Equation of the Directrix

The directrix is a horizontal line located \(p\) units from the vertex in the opposite direction of the parabola's opening. The vertex is at \[ (h, k) = (-4, 4) \] and since \[ k = 4, \] the directrix has the equation \[ y = k+p = 4+4 = 8. \]

Key Concepts

Vertex of a Parabola

The vertex of a parabola is one of its most essential features. Also known as the 'turning point,' it is where the parabola changes direction. In the context of the vertical parabola \((x+4)^2 = -16(y-4)\), the vertex can be identified by examining the standard vertex form of a parabola, which is \((x-h)^2 = 4p(y-k)\). This format reveals that the vertex is located at the point \((h, k)\).

For this equation, when compared to the standard form, \(h = -4\) and \(k = 4\). Thus, the vertex of the parabola is at \((-4, 4)\). This point represents the maximum or minimum value of the parabola depending on its orientation. Here, since the parabola opens downwards (due to the negative sign in front of the \(-4p\) term), the vertex at \((-4, 4)\) is its highest point.

Focus of a Parabola

The focus of a parabola is a unique point used to define its shape and location. Together with the directrix, it helps determine how the parabola 'bends.' The focus is located \(p\) units away from the vertex along the parabola’s axis of symmetry.

In our given equation \((x+4)^2 = -16(y-4)\), the term \(-16\) represents \(-4p\). Solving \(-4p = -16\) gives \(p = 4\). This value of \(p\) indicates the distance from the vertex to the focus.

Since this parabola opens downwards, the focus is located below the vertex. Thus, starting from the vertex \((-4, 4)\) and moving downward by \(4\) units (because \(p = 4\), and the parabola opens towards the negative \(y\)-direction), we find the focus at the point \((-4, 0)\). This means all points on the parabola are equidistant from this focus point and the directrix.

Directrix of a Parabola

The directrix of a parabola is a crucial line, integral in defining the parabola along with the focus. It is a horizontal line when dealing with vertical parabolas. The role of the directrix is to maintain a constant relationship with the focus, ensuring that each point of the parabola is equidistant to both the focus and the directrix.

For the parabola given by the equation \((x+4)^2 = -16(y-4)\), the directrix is located \(p\) units away from the vertex but in the opposite direction of the parabola's opening. Since \(p = 4\) and the parabola opens downwards, the directrix is above the vertex.

Starting from the vertex \((-4, 4)\), we move \(4\) units upward to find the directrix. This gives us the equation for the directrix: \(y = 8\). At this line, any point on the parabola has its distance to the focus and to the directrix equal, preserving the parabolic structure.