Problem 13 - College Algebra

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. $$ y=-4 x^{2} $$

Step-by-Step Solution

Answer

Vertex: (0, 0), Focus: \( (0, -\frac{1}{16}) \), Directrix: \( y = \frac{1}{16} \).

Step 1: Rewrite in Standard Form

The given equation is \( y = -4x^2 \). The standard form of a parabola that opens up or down is \( (x-h)^2 = 4p(y-k) \). We will convert the given equation to this form.Start by rearranging the terms:\[-4x^2 = y\] Divide both sides by -4:\[x^2 = -\frac{1}{4}y\] Rearrange to make it appear more like the standard form:\[x^2 = -\frac{1}{4}(y-0)\]In standard form, this represents \((x-h)^2 = 4p(y-k)\) with \(h=0\), \(k=0\), and \(4p = -\frac{1}{4}\). So, \(p = -\frac{1}{16}\).

Step 2: Identify the Vertex

From the standard form, \((x-h)^2 = 4p(y-k)\), we identify the vertex \((h, k)\) of the parabola. In our equation, \(h = 0\) and \(k = 0\).Therefore, the vertex is \(V = (0, 0)\).

Step 3: Determine the Focus

The distance from the vertex to the focus is \(p\). Since \(p = -\frac{1}{16}\) and the parabola opens downwards (negative \(p\)), the focus is \(p\) units below the vertex.Hence, the focus \((F)\) is at\[F = (0, 0 + p) = (0, -\frac{1}{16})\]

Step 4: Find the Directrix

The directrix of a parabola that opens upwards or downwards is a horizontal line at a distance \(p\) from the vertex, opposite to the direction of the parabola's opening.The parabola opens downwards, so the directrix is given by\[y = k - p = 0 + \frac{1}{16}\]Thus, the directrix is \(d: y = \frac{1}{16}\).

Key Concepts

Vertex of a Parabola

The vertex of a parabola serves as its turning point. In the parabolic equation's standard form, \((x-h)^2 = 4p(y-k)\), the coordinates of the vertex are \((h, k)\). In simple terms, picture the vertex as the parabola's center or peak for those opening upwards or downwards.

• In the given homework problem, the equation is initially \(y = -4x^2\).
• When rearranged into the standard form, we have \((x-0)^2 = -\frac{1}{4}(y-0)\).
• This readily provides the vertex coordinates \(V = (0, 0)\).

The vertex plays a critical role in graph interpretation by indicating where the parabola takes its minimum or maximum point, depending on its orientation.

Focus of a Parabola

The focus of a parabola is a special point that helps define the curve's shape. Every point on the parabola is equidistant from the focus and the directrix, an imaginary horizontal line parallel to the x-axis.

• In standard form, utilizing \((x-h)^2 = 4p(y-k)\), \(p\) is the distance from the vertex to the focus.
• For our example, \(p = -\frac{1}{16}\) indicates that the parabola opens downward.
• Hence, the focus \(F\) is located \(p\) units below the vertex: \((0, -\frac{1}{16})\).

Understanding the focus is crucial as it also impacts the parabola's reflective properties, such as how light or signals are focused to this point in real-world applications.

Directrix of a Parabola

The directrix is another fundamental element of a parabola that aids in defining its geometry. Unlike the focus, the directrix is a line, not a point. It serves as a reference for measuring equal distances from any point on the parabola.

• In the standard parabolic form, \((x-h)^2 = 4p(y-k)\), the directrix is represented by the line \(y = k - p\).
• For the problem at hand, with \(p = -\frac{1}{16}\) and \(k = 0\), the directrix is \(y = \frac{1}{16}\).
• This means all points along this line are used to maintain the parabola's constant width in relation to the focus.

Comprehending the concept of the directrix helps in grasping how parabolas visually and mathematically conform to their defined paths.