Problem 10 - Precalculus: Functions and Graphs

Question

Exer. \(1-12\) : Find the vertex, focus, and directrix of the parabola. Sketch its graph, showing the focus and the directrix. $$x^{2}+20 y=10$$

Step-by-Step Solution

Answer

Vertex: \((0, -\frac{1}{2})\), Focus: \((0, \frac{9}{2})\), Directrix: \(y = -\frac{11}{2}\).

Step 1: Rewrite Equation in Standard Parabola Form

We start with the given equation: \[x^2 + 20y = 10.\] To fit the standard parabola form \((x-h)^2 = 4p(y-k)\), let's solve for \(y\).Subtract \(10\) from both sides and divide by \(20\) to isolate \(y\): \[x^2 = 10 - 20y \Rightarrow 20y = x^2 - 10 \Rightarrow y = \frac{x^2}{20} - \frac{1}{2}.\] Which can be rewritten as: \[x^2 = 20\left(y + \frac{1}{2}\right).\]

Step 2: Identify Vertex, Focus, and Directrix

From the equation \(x^2 = 20(y + \frac{1}{2})\), we see it's of the form \((x-h)^2 = 4p(y-k)\) with \(h = 0\), \(k = -\frac{1}{2}\), and \(4p = 20\). Solving for \(p\), we find \(p = 5\). Thus, the vertex is at \((0, -\frac{1}{2})\).The focus of a parabola given this form is at \((h, k + p)\), i.e., \( (0, \frac{9}{2}) \).The directrix is the line \(y = k - p\), hence \(y = -\frac{11}{2}\).

Step 3: Sketch the Parabola and Mark Important Points

Sketching involves drawing the parabola opening upwards with vertex \((0, -\frac{1}{2})\). Mark the focus \((0, \frac{9}{2})\) and the directrix line \(y = -\frac{11}{2}\).Draw the curve equidistant from the focus and the directrix.

Key Concepts

Vertex of a Parabola

Understanding the vertex of a parabola is crucial as it marks the point where the parabola changes direction. In simpler terms, if a parabola opens upwards or downwards, the vertex is its lowest or highest point, respectively.
The vertex is given in the standard form equation of a parabola \[ (x-h)^2 = 4p(y-k) \] as the point \( (h, k) \). For the equation converted from the original exercise,\[ x^2 = 20(y + \frac{1}{2}) \], the vertex \((h, k)\) is located at \( (0, -\frac{1}{2}) \). This makes sense because at that point, the parabola is symmetrical and moves the same distance upwards on either side of the vertex.
To find the vertex:

• Rewrite the equation in its standard form. Here, \(x^2 = 20(y + \frac{1}{2}) \) shows that the vertex is at \( (0, -\frac{1}{2}) \).


• The vertex is the central theme around which a parabola is symmetric. Knowing the vertex helps plot and sketch the parabola effectively.

Focus of a Parabola

The focus is a specific point that helps define the parabola's shape and direction. It is one of the fundamental points in understanding parabolas as it reflects the distance a point on the curve maintains with the directrix.​
The focus is always situated "inside" the parabola towards the opening, distinct from the vertex.
In mathematical terms, the focus for our given parabola \([x^2 = 20(y + \frac{1}{2})] \) is calculated from the vertex formula. Use the formula:\[ ext{Focus} = (h, k + p) \]with \(h = 0, k = -\frac{1}{2}, p = 5 \)(from \(4p = 20\) which leads to \(p=5\)). Thus, it lies at \((0, \frac{9}{2})\). This spot helps in guiding the shape of the parabola.
Steps to find the focus:

• Identify the value \(p\) such that \(4p = 20\). This results in \(p = 5\).


• Apply this in the focus formula with the vertex values \((0, -\frac{1}{2})\) to result in \((0, \frac{9}{2})\).


• The focus always shifts the parabola into creating its characteristic 'U' shape enclosed around it.

Directrix of a Parabola

The directrix of a parabola is a straight line used alongside the focus to define the shape of the parabola. The distance between the directrix and any point on the parabola should equal the distance from the focus to the same point on the parabola.
Essentially, it, along with the focus, coaxes the parabola into its specific, symmetric form.
For the given parabola, the directrix can be derived using the formula:
\[ ext{Directrix} = y = k - p \]where \(h = 0, k = -\frac{1}{2}, p = 5\), leading to\[ y = -\frac{1}{2} - 5 = -\frac{11}{2} \], marking the directrix for this specific scenario.
Steps to find the directrix:

• Verify the equation maintains symmetry: given standard form implicates both focus and directrix working in tandem.


• Use the value \(k - p = -\frac{11}{2} \) to plot the directrix line.


• This complements the focus, keeping the curve balanced and authentic to its form.