Problem 4 - Calculus Early Transcendentals

Question

Find the vertex,focus, and directrix of the parabola and sketch its graph. $$y^{2}=12 x$$

Step-by-Step Solution

Answer

Vertex: (0,0), Focus: (3,0), Directrix: x=-3.

Step 1: Understanding the Standard Form

The given equation of the parabola is \(y^2 = 12x\). This is a horizontal parabola since the \(y\) term is squared and not the \(x\) term. The standard form of a horizontal parabola is \((y-k)^2 = 4p(x-h)\), where \((h, k)\) is the vertex and \(p\) is the distance from the vertex to the focus or the directrix.

Step 2: Identifying the Vertex

The given equation can be rewritten as \((y - 0)^2 = 12(x - 0)\). Comparing this with the standard form \((y-k)^2 = 4p(x-h)\), we identify \(h = 0\) and \(k = 0\). Therefore, the vertex of the parabola is \((0, 0)\).

Step 3: Finding the Value of \(p\)

From the standard form, \((y-k)^2 = 4p(x-h)\), we compare it with \((y - 0)^2 = 12(x - 0)\) to find \(4p = 12\). Solving for \(p\) gives \(p = 3\).

Step 4: Finding the Focus

For a horizontal parabola \((y-k)^2 = 4p(x-h)\), the focus is located at \((h+p, k)\). Substituting \(h = 0\), \(k = 0\), and \(p = 3\) into the equation for the focus gives \((0+3, 0) = (3, 0)\). Thus, the focus is \((3, 0)\).

Step 5: Finding the Directrix

The directrix of a horizontal parabola \((y-k)^2 = 4p(x-h)\) is the line \(x = h-p\). Substitute \(h = 0\) and \(p = 3\) to find \(x = 0 - 3 = -3\). Therefore, the directrix is \(x = -3\).

Step 6: Sketching the Parabola

To sketch the graph, plot the vertex at \((0, 0)\), the focus at \((3, 0)\), and draw the directrix as a vertical line \(x = -3\). Since it opens to the right due to the positive \(x\) on the right side of the equation, sketch a parabola that is symmetric about the \(x\)-axis, opening towards the focus.

Key Concepts

Vertex

The vertex of a parabola is the point that represents the highest or lowest point on the graph. In this specific exercise, we are dealing with a horizontal parabola, meaning the parabola opens to the right or left. A horizontal parabola takes the form \((y - k)^2 = 4p(x - h)\), where \((h, k)\) is the vertex.

• For our equation \(y^2 = 12x\), the vertex is at origin \((0, 0)\).
• The vertex provides a balanced point that helps in determining the direction of the parabola.

Understanding the vertex is crucial because other significant parts such as focus and directrix are related to this point. The vertex acts as the reference point from which you can measure distances for further steps in analysis. It serves as the 'starting position' when you sketch the parabola on a graph.

Focus

The focus of a parabola is a point located inside the curve, crucial for defining its shape. For a standard horizontal parabola given by \((y-k)^2 = 4p(x-h)\), the focus can be found at the point \((h+p, k)\).

• For our equation \(y^2 = 12x\), the value of \(p\) is calculated as 3, resulting in the focus being located at \((3, 0)\).
• The focus lies directly across from the vertex with respect to the direction the parabola opens, in our case, to the right.

The focus is important because all paths of the parabola reflect off towards this single point. A parabola's curve will always "bend" around this point, making it integral for understanding the geometry of parabolas in physics and engineering. It is the point towards which the parabola directs its opening, determining the set of points constituting the parabola's shape.

Directrix

The directrix of a parabola is a straight line that, together with the focus, helps to precisely describe the parabola. For a horizontal parabola, the directrix is given by the line \(x = h - p\). This line serves as a reference parallel to the \(y\)-axis, from which distances to the focus help determine the curve.

• In the equation \(y^2 = 12x\), since \(p = 3\), the directrix is the line \(x = -3\).
• It is equidistant from the focus on the opposite side of the vertex.

The significance of the directrix lies in its involvement in the parabola's reflection property: any point on the parabola is equidistant to both the focus and the directrix. This property is unique for parabolas and assists in accurately sketching the shape of the parabola when given the vertex and focus. Understanding the directrix allows you to capture the symmetry and balance of the parabola's graph.