Problem 20 - College Algebra with Corequisite Support

Question

For the following exercises, rewrite the given equation in standard form, and then determine the vertex \((V),\) focus \((F),\) and directrix \((d)\) of the parabola. \((x+1)^{2}=2(y+4)\)

Step-by-Step Solution

Answer

Vertex: \((-1, -4)\), Focus: \((-1, -\frac{7}{2})\), Directrix: \(y = -\frac{9}{2}\).

Step 1: Reformat the equation

The given equation \((x+1)^{2}=2(y+4)\)is in its vertex form. We want to express it in the standard form for a parabola which is \((x-h)^2=4p(y-k)\), where \((h,k)\) is the vertex of the parabola.

Step 2: Identify the Vertex

By comparing the given equation \((x+1)^2 = 2(y+4)\) with the standard form \((x-h)^2 = 4p(y-k)\),we can see that \(h = -1\)and \(k = -4\).Therefore, the vertex \(V\) is at \((h,k) = (-1,-4)\).

Step 3: Find the value of p

By comparing the coefficients of \((y-k)\), we have \(4p = 2\).Solving for \(p\), we find \(p = \frac{1}{2}\).

Step 4: Determine the Focus

The focus \(F\) of a vertical parabola is located \(p\) units away from the vertex along the axis of symmetry. The axis of symmetry is the line \(x = h\), so the focus is at \((h, k + p)\).Thus the focus \(F\) is at \((-1, -4 + \frac{1}{2}) = (-1, -\frac{7}{2})\).

Step 5: Find the Directrix

The directrix \(d\) of a vertical parabola is a horizontal line \(p\) units away from the vertex, opposite the focus. For our equation, this line is \(y = k - p\).The directrix is: \(y = -4 - \frac{1}{2} = -\frac{9}{2}\).

Key Concepts

Vertex

In the world of parabolas, the vertex is a core feature. It represents the turning point and the location of the parabola's minimum or maximum point. For the equation \((x+1)^{2}=2(y+4)\), we can derive the vertex by expressing the equation in vertex form \((x-h)^2=4p(y-k)\). Here, \(h\) and \(k\) define the coordinates of the vertex. When we compare our equation with the standard vertex form, it's clear that \(h = -1\) and \(k = -4\). Therefore, the vertex is located at the point \((-1, -4)\).

• The vertex is the point where the parabola changes direction.
• For upward or downward opening parabolas, it serves as the minimum or maximum point.

Understanding the vertex helps in graphing the parabola, as it gives a central reference point from which other features radiate.

Focus

The focus of a parabola is a significant point located inside the curve. It's an essential part of the parabola's reflective property. Every point on the parabola is equidistant from the focus and the directrix, creating the curve.

For our example \((x+1)^{2}=2(y+4)\), we discovered that the parabola is vertical. To find the focus, we use the vertex and the value \(p\), which represents the distance from the vertex to the focus. With \(p = \frac{1}{2}\), the focus is located \(\frac{1}{2}\) units away from the vertex along the line \(x=-1\). Thus, the focus is at the point \((-1, -\frac{7}{2})\).

• The focus is always located on the inside of the parabola.
• It works alongside the directrix to define the parabola uniquely.

The concept of focus is fundamental in applications around optics and satellite dishes, where reflective properties of parabolas are utilized.

Directrix

Opposite to the focus, the directrix helps shape the parabola, lying in a straight line away from it. The directrix for a vertical parabola is a horizontal line. In our case, the equation \((x+1)^{2}=2(y+4)\) allows us to identify the directrix by moving \(p\) units away from the vertex in the opposite direction of the focus.

Thus, for the vertex \((-1, -4)\) and \(p = \frac{1}{2}\), the directrix is the line \(y = -4 - \frac{1}{2} = -\frac{9}{2}\).

• The directrix is a fixed line used to control the curvature of the parabola.
• It's a critical element in verifying each point belongs to the parabola.

The concept of the directrix, alongside the focus, makes it possible to construct parabolas with specific properties for engineering and physics applications.