Problem 6 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-12: Find the vertex, focus, and directrix of the parabola. Sketch its graph, showing the focus and the directrix. $$ (x-3)^{2}=\frac{1}{2}(y+1) $$

Step-by-Step Solution

Answer

Vertex: \((3, -1)\), Focus: \((3, -\frac{7}{8})\), Directrix: \(y = -\frac{9}{8}\).

Step 1: Identify the Parabola Form

The equation \((x-3)^2 = \frac{1}{2}(y+1)\) is in the standard form \((x-h)^2 = 4p(y-k)\), where \((h, k)\) is the vertex and \(p\) is the distance from the vertex to the focus and the directrix.

Step 2: Determine the Vertex

The given equation \((x-3)^2 = \frac{1}{2}(y+1)\) shows the vertex form \((x-h)^2 = 4p(y-k)\). Here \(h = 3\) and \(k = -1\). Thus, the vertex is at \((3, -1)\).

Step 3: Solve for 4p

Compare \(\frac{1}{2}\) with \(4p\) to find \(p\). We have \(4p = \frac{1}{2}\), so \(p = \frac{1}{8}\).

Step 4: Find the Focus

The focus of the parabola is at \((h, k + p)\). Substitute the vertex \((3, -1)\) and \(p = \frac{1}{8}\) to get \((3, -1 + \frac{1}{8}) = (3, -\frac{7}{8})\).

Step 5: Determine the Directrix

The directrix is the line \(y = k - p\). Using \(k = -1\) and \(p = \frac{1}{8}\), the directrix is \(y = -1 - \frac{1}{8} = -\frac{9}{8}\).

Step 6: Sketch the Graph

The parabola opens upwards since the equation is derived from \((x-h)^2 = 4p(y-k)\) with a positive \(p\). Draw the vertex at \((3, -1)\), the focus at \((3, -\frac{7}{8})\), and the directrix as the horizontal line \(y = -\frac{9}{8}\). Make sure to indicate the direction of opening and label the focus and directrix on your sketch.

Key Concepts

Vertex of a Parabola

The vertex of a parabola is a key point, marking the spot where the parabola turns. Think of it as the top of a hill or bottom of a valley. In mathematical terms, the vertex is represented by the coordinates \((h, k)\). These values provide a specific location on the coordinate plane where the parabola's axis of symmetry intersects the curve. For example, in our equation \((x-3)^2 = \frac{1}{2}(y+1)\), the vertex is at \((3, -1)\).
The vertex is important because it acts as a reference point for the rest of the parabola. From the vertex, the curve extends symmetrically in two opposite directions. In an upward-facing parabola like ours, the vertex is the lowest point. For a downward-facing parabola, it would be the highest point.
Finding the vertex early on makes it simpler to analyze the parabola's other features, such as the focus and directrix, as they are all calculated relative to the vertex.

Focus of a Parabola

The focus of a parabola is another crucial component that helps define its shape. The focus is a point located inside the curve, and it serves as a point that every point on the parabola is equidistant from, in relation to the directrix. In our equation, the focus can be found by looking at the distance \(p\) from the vertex. This is derived from the equation \(4p(y-k)\), where \(p\) is the factor determining how 'steep' or 'wide' the parabola is.

• In our example, \(p = \frac{1}{8}\), calculated by setting \(4p = \frac{1}{2}\).
• The focus is located at \((h, k + p)\), rendering it \( (3, -\frac{7}{8})\) for our specific problem.

This location is vital when sketching the graph of the parabola. It provides orientation, indicating where inside the parabola the curve is "aiming" at or "bending" towards.

Directrix of a Parabola

The directrix of a parabola is a line that, together with the focus, helps define the parabola’s curve. Unlike the focus, which is a point, the directrix is a straight line from which distances are equated to determine the points on the parabola. The line runs parallel to the axis of symmetry and perpendicular to the direction in which the parabola opens.
The mathematical role of the directrix is to keep the curve balanced. For any point \((x, y)\) on the parabola, the distance to the focus and the distance to the directrix is equal. In our equation, the directrix is the horizontal line located at \(y = k - p\). With \(k = -1\) and \(p = \frac{1}{8}\), the directrix lies at \(-\frac{9}{8}\).
The importance of the directrix lies in helping visualize and sketch the parabola accurately. By marking the directrix on a graph, one can support the correct plotting of the curve, ensuring it maintains its defined distance from both the focus and the line.