Problem 5 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Vertex: (-2, 3); Focus: (-2, 5); Directrix: y = 1; The parabola opens upwards.

Step 1: Convert the Equation into Standard Form

The equation given is \((x+2)^{2} = 8(y-3)\). This is already in the standard form of a parabola: \((x-h)^2 = 4p(y-k)\), where \(h, k\) is the vertex and \(p\) is the focal length. This indicates that the parabola opens vertically. Here, \(h = -2\) and \(k = 3\), so the vertex is at \((-2, 3)\).

Step 2: Identify the Vertex

From the form \((x-h)^2 = 4p(y-k)\), the vertex is easily identified as the point \((h, k)\). Hence, the vertex of the given parabola is \((-2, 3)\).

Step 3: Determine 'p' and the orientation of the Parabola

In the equation \((x+2)^{2} = 8(y-3)\), compare it with \((x-h)^2 = 4p(y-k)\). We see that \(4p = 8\), hence \(p = 2\). Since the equation is in the form \((x-h)^2 = 4p(y-k)\), the parabola opens upwards if \(p > 0\).

Step 4: Find the Focus

The focus of a parabola that opens upwards \((x-h)^2 = 4p(y-k)\) is located at \((h, k + p)\). With \(h = -2\) and \(k = 3\), and \(p = 2\), the focus is at \((-2, 3 + 2) = (-2, 5)\).

Step 5: Determine the Directrix

The directrix of a parabola that opens upwards is a horizontal line given by \(y = k - p\). Substituting \(k = 3\) and \(p = 2\), the equation for the directrix is \(y = 3 - 2 = 1\).

Step 6: Sketch the Graph

To sketch the graph, plot the vertex at \((-2, 3)\), the focus at \((-2, 5)\), and draw the directrix at \(y = 1\). Since \(p\) is positive, the parabola opens upwards, symmetrically about the axis of symmetry \(x = -2\). Draw the parabola opening upwards from the vertex towards the focus.

Key Concepts

The Vertex of a Parabola

The vertex of a parabola is a key feature. It can be thought of as the "tip" or the peak, depending on how the parabola opens. For a parabola described by the equation \((x-h)^2 = 4p(y-k)\), the vertex is located at \((h, k)\). This is because \((h, k)\) represents the point from which the parabola is exactly symmetrical. If you think of a parabola as a "U" shape, \((h, k)\) is the lowest point for an upwards-facing parabola and the highest point for a downwards-facing one.

• The vertex provides a starting point for sketching a parabola.
• It helps in determining the direction the parabola faces.

For example, in the equation \((x+2)^2 = 8(y-3)\), the vertex can be found by identifying \(h\) and \(k\). Adjust your equation into the form \((x-h)^2 = 4p(y-k)\), where \(h = -2\) and \(k = 3\).Therefore, the vertex is at the point \((-2, 3)\), which is crucial for graph plotting.

The Focus of a Parabola

The focus of a parabola is a special point that, along with the directrix, helps define the shape of the parabola. It is located inside the curve of the parabola. In our typical parabola formula \((x-h)^2 = 4p(y-k)\), the focus is found at the point \((h, k + p)\) for a parabola that opens upwards or downwards.

• For a left or right opening parabola, the location would differ to \((h + p, k)\).
• Your specific parabola's focal point will always be \(p\) units away from the vertex, measured along the axis of symmetry.

In the example \((x+2)^2 = 8(y-3)\), finding \(p\), which equals\(2\), allows us to find the focus at\((-2, 5)\). Knowing the focus is important because it helps in understanding how the parabola directs its curves towards that point.

Understanding the Directrix

The directrix of a parabola complements the focus by being a line that acts as a reference for measuring distances from any point on the parabola backwards toward this line. In simple terms, every point on a parabola is equidistant from the directrix and the focus.

• The directrix is always perpendicular to the axis of symmetry of the parabola.
• Just like the focus, it helps to fine-tune the shape and width of the parabola.

For the parabola \((x-h)^2 = 4p(y-k)\), if it opens upward or downward, the directrix is a horizontal line given by \(y = k - p\). In our specific problem \((x+2)^2 = 8(y-3)\), where \(k = 3\) and \(p = 2\), we find the directrix at \(y = 1\). Recognizing the directrix is useful whenever plotting or analyzing parabola behavior since it provides a boundary measure relative to the vertex and focus.