Problem 5 - Algebra and Trigonometry

Question

Find the vertex, focus, and directrix of the parabola, and sketch the graph. $$ (x-3)^{2}=8(y+1) $$

Step-by-Step Solution

Answer

Vertex: (3, -1); Focus: (3, 1); Directrix: y = -3.

Step 1: Identify the Parabola's Standard Form

The given parabola equation \((x-3)^2 = 8(y+1)\) is in the form \((x-h)^2 = 4p(y-k)\), which represents a vertical parabola. Here, the vertex is \((h, k)\).

Step 2: Determine the Vertex

From the equation \((x-3)^2 = 8(y+1)\), compare it with the standard form \((x-h)^2 = 4p(y-k)\). We have \(h = 3\) and \(k = -1\). Thus, the vertex of the parabola is \((3, -1)\).

Step 3: Calculate the Value of p

In the standard form \((x-h)^2 = 4p(y-k)\), we equate \(4p = 8\). Solving for \(p\), we find \(p = 2\).

Step 4: Find the Focus

The focus is located at \((h, k + p)\). With \(h = 3\), \(k = -1\), and \(p = 2\), the focus is \((3, 1)\).

Step 5: Determine the Directrix

The directrix of a vertical parabola \((x-h)^2 = 4p(y-k)\) is given by \(y = k - p\). Plugging in the values, \(y = -1 - 2 = -3\), so the directrix is \(y = -3\).

Step 6: Sketch the Graph

The vertex is at \((3, -1)\), the focus at \((3, 1)\), and the directrix at \(y = -3\). The parabola opens upwards (since \(p > 0\)). Plot the vertex and focus, draw the directrix as a horizontal line, and sketch the parabola opening upwards.

Key Concepts

Vertex

The vertex of a parabola is a key point where the parabola changes direction. In the equation \[(x-3)^2 = 8(y+1),\]we identify the vertex by comparing it to the standard form of a vertical parabola: \[(x-h)^2 = 4p(y-k).\]

• Here, \(h = 3\) and \(k = -1\).
• Thus, the vertex is located at \((3, -1)\).

The vertex represents either the highest or lowest point on the graph of a parabola. For this specific parabola, it is the lowest point given that it opens upwards. The coordinates of the vertex are crucial for sketching the parabola and understanding its orientation.

Focus

The focus of a parabola is a point inside the curve that helps in defining its shape and direction. From the equation \((x-3)^2 = 8(y+1)\), we've already determined that \(p=2\), which helps find the focus.

• The focus for a vertical parabola can be found using the formula \((h, k+p)\).
• Substituting the values: \(h = 3\), \(k = -1\), and \(p = 2\), we find the focus to be \((3, 1)\).

The focus is essential because it lies along the axis of symmetry of the parabola. Rays of light or other signals originating from the focus will reflect off the parabola and travel in specific directions, which is why this point is of great significance in applications such as satellite dishes and headlights.

Directrix

The directrix is a line perpendicular to the axis of symmetry of the parabola and plays an essential role in its geometric definition. For the parabola \[(x-3)^2 = 8(y+1),\]the equation of the directrix can be found by using the formula for a vertical parabola: \[y = k - p,\]where \(k = -1\) and \(p = 2\).

• Thus, the directrix is \(y = -1 - 2 = -3\).

The directrix is always located opposite the focus relative to the vertex, forming one part of the key geometry that defines a parabola. It contributes to the unique property of parabolas: the distance from any point on the parabola to the focus is equal to the perpendicular distance from the point to the directrix, a property that is fundamental to understanding the parabola's geometry and behavior.