Problem 61 - Algebra 2

Question

Identify the focus and directrix of each parabola. Then graph the parabola. $$ x=-\frac{1}{4} y^{2} $$

Step-by-Step Solution

Answer

The focus of the parabola is \((-1/4, 0)\) and the directrix is the line \(x = 1/4\). The graph of the parabola opens to the left with the vertex at the origin.

Step 1: Identifying 'a' from the equation

The equation of the parabola is given in the form \(x = ay^2\). Comparing this to the given equation \(x = -1/4y^2\), we can identify that \(a = -1/4\).

Step 2: Finding the focus

The focus of a parabola given in the form \(x = ay^2\) is located at \((a,0)\). Substituting the identified value of \(a\) from step 1, we find that the focus is \((-1/4, 0)\).

Step 3: Finding the directrix

The directrix of a parabola given in the form \(x = ay^2\) is the line \(x = -a\). Substituting the identified value of \(a\) from step 1, we find that the equation of the directrix is \(x = 1/4\).

Step 4: Graphing the parabola

Draw the graph taking into account that the vertex is at the origin (0, 0), the focus is at (-1/4, 0) and the directrix is \(x=1/4\). Draw the parabola such that it opens towards the left as \(a\) is negative and it wraps around the focus and is adjacent to the directrix.

Key Concepts

Focus

The focus of a parabola is an important point. It lies inside the curve of the parabola and is one point you need to clearly locate when graphing a parabola. For parabolas that open left, right, upward, or downward, the focus is always inside the parabola. So, it's never on the arms or outside of the curve.
For our specific equation, given in the form \(x = ay^2\), the focus is at \((a, 0)\). Here, we found \(a = -1/4\), so the focus is at \((-1/4, 0)\). Placing this on a graph helps determine the entire shape of the parabola. Keep in mind how this focus affects the direction and width of the parabola as well.

Directrix

The directrix of a parabola is a crucial element. It's a fixed line, which together with the focus, helps define the parabola. For a parabola opening sideways, like ours, the directrix won't intersect the parabola itself but helps maintain the symmetry.
For our equation, in the format \(x = ay^2\), the directrix is given by the line \(x = -a\). We calculated \(a = -1/4\), making the directrix line \(x = 1/4\). This line is always at the same distance from the vertex as the focus is, just in opposite directions.
When you plot it on the graph, you ensure your parabola is correctly positioned between the directrix and the focus.

Vertex

The vertex is another key feature of parabolas. It's the point on the curve that is either at the "tip" or "center" of the parabola, depending on its orientation. It's the halfway point between the focus and the directrix.
In our case, where the parabola is in the form \(x = ay^2\), the vertex conveniently is located at the origin, \((0,0)\). The location of the vertex can change in other forms of equations, but here it's fixed and straightforward.
The vertex acts as the point of symmetry for the parabola, which helps ensure the shape and position of the graph are accurate on the coordinate plane.

Graphing

Graphing a parabola involves understanding its components like the focus, directrix, and vertex, and how they interact. Our equation, \(x = -\frac{1}{4} y^2\), demonstrates a sideways-opening parabola, which is unique because it has the vertex at the origin \( (0,0) \).
Here’s how to approach graphing this parabola:

• Start by plotting the vertex at \((0,0)\).
• Next, mark the focus at \((-1/4,0)\).
• Line up the directrix to \(x = 1/4\).
• Now, sketch your parabola so that it opens left towards the focus.

Ensure the arms or branches of the parabola smoothly curve around the focus and stay in the region between the vertex and approaching the directrix but never touching it. This relationship will portray the parabolic shape demanded by the equation.