Problem 5 - Precalculus: Mathematical for Calculus

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 standard form of the parabola

The given equation \( (x-3)^2 = 8(y+1) \) is of the form \( (x-h)^2 = 4p(y-k) \), where \( (h, k) \) is the vertex. Comparing \(8(y+1)\) to \(4p(y-k)\), we identify that \(4p = 8\). Thus, \(p = 2\).

Step 2: Determine the vertex

From the given equation \((x-3)^2 = 8(y+1)\), we see that \(h = 3\) and \(k = -1\). Thus, the vertex of the parabola is \((3, -1)\).

Step 3: Find the focus

The focus of the parabola \((x-h)^2 = 4p(y-k)\) is given by \((h, k+p)\). Substituting \(h = 3\), \(k = -1\), and \(p = 2\), we get the focus as \((3, -1 + 2) = (3, 1)\).

Step 4: Determine the equation of the directrix

The directrix of a parabola in the form \((x-h)^2 = 4p(y-k)\) is \(y = k-p\). With \(k = -1\) and \(p = 2\), the directrix equation becomes \(y = -1 - 2 = -3\).

Step 5: Sketch the graph

The parabola opens upwards because the coefficient of \(y\) is positive. Plot the vertex at \((3, -1)\) and the focus at \((3, 1)\). Draw the directrix as \(y = -3\). Ensure the opening of the parabola points in the direction of the focus.

Key Concepts

Vertex

In a parabola, the vertex is a crucial point that indicates where the parabola changes direction. It's the "turning point" of the curve. For our given parabola equation

• \((x-3)^2 = 8(y+1)\),
• The vertex form is \((x-h)^2 = 4p(y-k)\).

This shows that the vertex is located at \((h, k)\). In our equation, \(h = 3\) and \(k = -1\), so the vertex is at \((3, -1)\).
The vertex is important because it helps in sketching the parabola and finding other features such as the axis of symmetry.

• The vertex for upward or downward parabolas is the point where the curve reaches its minimum or maximum.

With the vertex known, you can easily begin sketching the parabola and see how it would look on a graph.

Focus

The focus of a parabola is another significant point. It lies inside the curve and, together with the directrix, helps define the shape of the parabola. Simply put, a parabola is the set of all points that are equidistant from the focus and the directrix. In the case of our parabola

• \((x-3)^2 = 8(y+1)\),
• The focus is computed using the form \((h, k+p)\),

where \(p\) is known from the equation \(4p(y-k)\). With \(h = 3\), \(k = -1\), and \(p = 2\), the focus lies at \((3, 1)\).
The focus helps determine the direction in which the parabola opens.

• If the parabola opens upwards, the focus sits above the vertex.
• If the parabola opens downwards, the focus is below the vertex.

For our parabola, since the focus is at \((3, 1)\) and the parabola opens upwards, it confirms the upward direction of the opening.

Directrix

The directrix is an essential feature of the parabola that works along with the focus. It serves as a line reference point. Every point on the parabola is equidistant from the focus and the directrix.
For the given equation

• \((x-3)^2 = 8(y+1)\),
• We use the standard form \((x-h)^2 = 4p(y-k)\)

to determine the directrix. Using the value of \(k\) and \(p\), the directrix can be found with the formula \(y = k-p\). With \(k = -1\) and \(p = 2\), the directrix is \(y = -3\).

• The directrix is always perpendicular to the axis of symmetry.

In our parabola, the directrix \(y = -3\) gives us a fixed line below the vertex. It helps ensure that the entire structure is symmetrical and balanced with the focus.