Problem 9 - Precalculus Mathematics for Calculus

Question

Find the vertex, focus, and directrix of the parabola. Then 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 form of the parabola

The given equation \((x-3)^{2}=8(y+1)\) resembles the standard form of a vertical parabola equation \((x-h)^{2}=4p(y-k)\). Here, the vertex form is evident where \(h=3\) and \(k=-1\). Therefore, this parabola has a vertex at \((h, k)\).

Step 2: Determine the vertex

From the equation \((x-3)^{2}=8(y+1)\), we can see that \(h = 3\) and \(k = -1\). Therefore, the vertex is at the point \((3, -1)\).

Step 3: Calculate the value of 4p

The equation in the form \((x-h)^{2}=4p(y-k)\) shows that \(4p = 8\). Therefore, we can solve for \(p\) by dividing both sides by 4 to get \(p = 2\).

Step 4: Identify the focus

The focus of a parabola defined by \((x-h)^{2}=4p(y-k)\) is at \((h, k+p)\). Substituting \(h = 3\), \(k = -1\), and \(p = 2\), we find the focus to be at \((3, 1)\).

Step 5: Determine the directrix

The directrix of a vertical parabola \((x-h)^{2}=4p(y-k)\) is given by the equation \(y = k - p\). Substituting \(k = -1\) and \(p = 2\), the equation of the directrix is \(y = -3\).

Step 6: Sketch the graph

With the vertex at \((3, -1)\), the focus at \((3, 1)\), and the directrix \(y = -3\), sketch a vertical parabola opening upwards. The vertex is the parabola's lowest point, and it will go upwards, passing through the focus and opening away from the directrix.

Key Concepts

Vertex of a Parabola

The vertex of a parabola represents a crucial point where the curvature is the highest or lowest on its graph. To find it, you look into the equation of the parabola. In standard form, a parabola can be either vertical or horizontal. Here, the equation

• (x-3)\(^2\) = 8(y+1)

indicates a vertical parabola, easily fit into the standard vertex form

• (x-h)\(^2\)=4p(y-k)
.

This particular format explicitly provides the location of the vertex, denoted by the coordinates

• (h, k).

In this case, the vertex is found by identifying the values of h and k from the equation, which are 3 and -1 respectively. Thus, the vertex is at the point (3, -1).
The vertex is primarily responsible for defining the direction and general shape of the parabola on the graph. For vertical parabolas, such as this one, the vertex is the "turning point" from which the curve rises or falls.

Focus of a Parabola

The focus is another critical element in understanding the geometry and behavior of parabolas. For a parabola in the form

• (x-h)\(^2\) = 4p(y-k)

, you can determine the focus by using the formula

• (h, k+p).

It is always located on the axis of symmetry of a parabola and is a fixed point that helps guide the shape of the parabola's curve.
Looking at our equation, we have h = 3, k = -1 and we deduced that p = 2. Thus, plugging these values into the focus formula, the focus of this parabola is at the coordinates (3, 1).
The focus is a point where all the light that comes parallel to the axis of symmetry converges. It might feel a bit abstract, but it's a central part of understanding how the parabola functions in a practical context, such as in satellite dishes or headlights of cars.

Directrix of a Parabola

The directrix of a parabola is a straight line that interacts with the focus to define the parabola's shape. Unlike the focus, which is a point, the directrix is a line parallel to the axis of the parabola's opening.
For vertically oriented parabolas, as represented by the equation

• (x-h)\(^2\) = 4p(y-k)

,the formula to determine the directrix is

• y = k - p.

Using the values from our example, k = -1 and p = 2, we find the directrix to be aligned with the line y = -3.
While the focus attracts points on the parabola, the directrix serves to balance this attraction by establishing a fixed distance measure, known as the focal length, from any point on the parabola to itself and the focus. The directrix is essential in defining the precise shape and location of the parabola by maintaining equilibrium with the focus along the curve's expanse.