Problem 1 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-12: Find the vertex, focus, and directrix of the parabola. Sketch its graph, showing the focus and the directrix. $$ 8 y=x^{2} $$

Step-by-Step Solution

Answer

The vertex is (0, 0), focus is (0, 2), and the directrix is y = -2.

Step 1: Write the Parabola Equation in Standard Form

The given equation is \( 8y = x^2 \). This is a vertical parabola in standard form \( x^2 = 4py \). To rewrite it, divide both sides by 8 to get \( y = \frac{1}{8}x^2 \). The equivalent form \( x^2 = 8y \) reveals it's already in standard form with \( 4p = 8 \). Here, \( p = 2 \).

Step 2: Identify the Vertex

For the equation \( x^2 = 4py \), the vertex is at the origin \( (0, 0) \).

Step 3: Determine the Focus

The focus of a parabola \( x^2 = 4py \) is located at \( (0, p) \). With \( p = 2 \), the focus is at \( (0, 2) \).

Step 4: Find the Directrix

The directrix of a parabola \( x^2 = 4py \) is at \( y = -p \). Therefore, with \( p = 2 \), the directrix is \( y = -2 \).

Step 5: Sketch the Parabola

Draw the parabola with vertex at the origin. It opens upwards since it's in the form \( x^2 = 4py \) with \( p > 0 \). Mark the focus at \( (0, 2) \) and the directrix at \( y = -2 \).

Key Concepts

Understanding the Vertex

The vertex of a parabola is a key point that represents the lowest or highest point of the curve, depending on its orientation. In the equation \[x^2 = 4py,\]the vertex lies at the point (0,0) if there are no shifts or modifications in the equation. This is the case for our parabola \(8y = x^2\), which translates to \[y = \frac{1}{8}x^2.\]

• The vertex, therefore, is at the origin \((0,0)\).
• This point signifies where the parabola changes direction, creating a smooth "U" shape (if opening upwards) or an inverted "U" (if opening downwards).
• For vertical parabolas like ours, the vertex is the minimum point.

To find it in an equation, you can usually locate the vertex by solving for where the derivative (rate of change) of the function equals zero or directly from standard form, emphasizing simplicity.

Exploring the Focus

The focus of a parabola is an essential feature, being a point off the vertex used alongside the directrix to define the parabola's shape. For our specific equation in the form \(x^2 = 4py\),

• The parameter \(p\) translates to the distance from the vertex to the focus as \(p\) units.
• In our parabola, \(p = 2\), placing the focus at the point \((0,2)\).
• This placement indicates that as the parabola opens upwards, it maintains this distance before touching an imaginary horizontal line at the focus.

The focus is crucial as every point on the parabola is equidistant from this point and the directrix, forming part of the geometric principle defining parabolas.

Locating the Directrix

In a parabola, the directrix is a straight line that, combined with the focus, helps shape the parabola. The directrix serves as an equal balance with the focus, lying at an equal yet opposite distance from each point on the parabola.

• For the equation \(x^2 = 4py\), the directrix is a horizontal line given by \(y = -p\).
• In our example, \(p = 2\), so the directrix is \(y = -2\).
• This line is crucial in balancing the parabola, which bows towards the focus, providing a baseline for its curve.

Understanding the directrix helps visualize the parabola as part of a broader principle where each point on the curve remains equidistant between the focus and this line.

Graph Sketching of Parabolas

Graphing a parabola involves integrating the knowledge of its vertex, focus, and directrix. It's like creating a blueprint of a balanced curve.

• Begin by marking the vertex at \((0,0)\), the starting point from which the parabola spreads out.
• Identify the focus at \((0,2)\) and draw a small dot to represent it.
• Next, draw the directrix as a horizontal line at \(y = -2\).
• Sketch the parabola opening upwards, ensuring it lies equidistant between the focus and directrix.

This process not only visualizes the equation but also enhances comprehension of how varying \(p\) values affect the parabola's width and direction. By sketching, students solidify their understanding of the parabolic shape and its defining characteristics.