Problem 40 - Precalculus

Question

Find the vertex, focus, and directrix of the parabola. Then sketch the parabola. $$x=\frac{1}{4}\left(y^{2}+2 y+33\right)$$

Step-by-Step Solution

Answer

The vertex of the parabola is (1,32), the focus is (1,33) and the directrix is \(y=31\). The parabola opens upward.

Step 1: Rewriting the equation into standard form

To begin with the process, the given equation has to be rewritten into a standard form of parabola \(y=ax^2+bx+c\), where vertex \((h,k)\) is given by \(-\frac{b}{2a}\), \(c-\frac{b^2}{4a}\). Completing the square, the given equation can be rewritten as \(x=\frac{1}{4}(y-1)^2+32\).

Step 2: Identifying the Vertex

From the standard form, the vertex \((h,k)\) can be found. In this case, \(h\) equals 1 and \(k\) equals 32. Therefore, the vertex of the parabola is \((1,32)\).

Step 3: Identifying the Focus

The focus of the parabola can be found using the formula \((h, k+ \frac{1}{4a})\), where \(a\) is the coefficient of \(y^2\). In this case, it is \(\frac{1}{4}\). Thus the focus is \((1, 32+\frac{1}{4*1/4})\) equals to \((1,33)\).

Step 4: Identifying the Directrix

The formula for the directrix is \(y = k - \frac{1}{4a}\). Substituting values, \(y = 32 - \frac{1}{4*\frac{1}{4}} = 31\)

Step 5: Sketching the Parabola

With the vertex at (1, 32), the focus at (1,33) and the directrix \(y=31\), the Parabola opens upwards. First, plot the vertex. Then draw a vertical line at \(y=31\) as the directrix. Next plot the focus. Then sketch the parabola such that it is wrapped around the focus and parallel to the directrix.

Key Concepts

Vertex

The vertex is a key point on the parabola. It is where the curve changes direction. In simpler terms, you can think of it as the peak or the trough of the parabola.
To find the vertex, we start by getting the equation into its standard form. For our parabola equation, \[ x = \frac{1}{4}(y-1)^2 + 32 \] this is in the standard form of \[ x = a(y-k)^2 + h \].
Here, \(h\) and \(k\) directly represent the vertex coordinates. For our problem, \(h = 1\) and \(k = 32\), making the vertex \((1, 32)\).

• It is central to the parabola's shape.
• Acts as the point that is equidistant from both the focus and the directrix.

Understanding its role helps in plotting and visualizing the parabola effectively.

Focus

The focus is a significant component that helps define a parabola's shape. It is located inside the parabola and directly influences how wide or narrow the parabola opens.
The equation for the focus is often derived from modifying the vertex form of the parabola equation. In the given equation,\(a = \frac{1}{4}\), which was used to find the focus point. In our example, we found the focus to be at \((1, 33)\).

The focus determines:

• The direction in which the parabola opens (here it opens upwards).
• The distance to any point and its reflection on the other side of the directrix.

These reflections are equidistant from the focus, helping to ensure the symmetric shape of the parabola.

Directrix

The directrix is an essential line that is perpendicular to the axis of the parabola, contributing to its symmetric properties.
In our scenario, the directrix formula is used to find the line \(y = 31\).
Here's how it breaks down:

• Use the formula for the directrix, \(y = k - \frac{1}{4a}\), with \(a\) as the coefficient of \(y^2\), which is found to be \(\frac{1}{4}\) in our given problem.
• This directrix lies opposite to the opening direction of the parabola.
• It's important because each point on the parabola is equidistant to the focus and directrix.

The directrix helps to ensure that the curve maintains its unique parabolic shape, thus playing a critical role in understanding and graphing the parabola.