Problem 10 - Thomas Calculus

Question

Exercises \(9-16\) give equations of parabolas. Find each parabola's focus and directrix. Then sketch the parabola. Include the focus and directrix in your sketch. $$ x^{2}=6 y $$

Step-by-Step Solution

Answer

Focus: \( (0, \frac{3}{2}) \); Directrix: \( y = -\frac{3}{2} \).

Step 1: Identify the Standard Form of the Parabola

The given equation is \(x^2 = 6y\). This equation resembles the standard form of a parabola that opens vertically: \((x - h)^2 = 4p(y - k)\), where \((h, k)\) is the vertex. In this case, the vertex \((h, k)\) is at the origin \((0, 0)\).

Step 2: Determine the Value of \(p\)

In the equation \(x^2 = 6y\), we can compare it with \((x - 0)^2 = 4p(y - 0)\), which reveals \(4p = 6\). Solving for \(p\), \(p = \frac{6}{4} = \frac{3}{2}\).

Step 3: Locate the Focus

The focus of the parabola is \((h, k + p)\). Since \(h = 0\), \(k = 0\), and \(p = \frac{3}{2}\), the focus is at \(\left(0, \frac{3}{2}\right)\).

Step 4: Find the Directrix

The directrix of the parabola is \(y = k - p\). Substituting \(k = 0\) and \(p = \frac{3}{2}\), the directrix is \(y = -\frac{3}{2}\).

Step 5: Sketch the Parabola

To sketch the parabola, place the vertex at the origin \((0,0)\). The parabola opens upwards along the y-axis. Mark the focus at \((0, \frac{3}{2})\) and draw the directrix as a horizontal line at \(y = -\frac{3}{2}\). The parabola will be symmetric around the y-axis and pass through the focus as per its definition.

Key Concepts

Focus of a parabola

Every parabola has a special point known as its *focus*, which is crucial for defining its shape. The focus lies inside the parabola, and the parabola is shaped such that any point on it is equidistant from the focus and the directrix. To find the focus of a parabola, we use its standard form. For a parabola like the one in the exercise, given by the equation \(x^2 = 6y\), which is similar to the form \((x-h)^2 = 4p(y-k)\), the focus can be identified.In our case, the vertex \((h, k)\) is \((0, 0)\), and \(p = \frac{3}{2}\) as calculated. The focus, therefore, is located at \((0, \frac{3}{2})\), directly above the vertex. While sketching, remember that the parabola curves around this point, making it an essential guide for sketching accurately.

Directrix of a parabola

The directrix of a parabola is a line that plays a vital role alongside the focus in defining the curve's shape. The points on the parabola are equidistant from both the focus and this line. In mathematical terms, for a vertically oriented parabola like \(x^2 = 6y\), the directrix line can be determined using the standard form equation.For our specific example, with the vertex at the origin \((0, 0)\) and \(p = \frac{3}{2}\), the directrix is the line \(y = k - p\), which simplifies to \(y = -\frac{3}{2}\). When sketching the parabola, the directrix helps in forming the pathway of the curve. It's located directly opposite the focus from the vertex, providing symmetry essential for an accurate sketch.

Sketching a parabola

Sketching a parabola involves plotting crucial components such as the vertex, focus, and directrix, which guide the shape. Begin by determining these elements, as done in the exercise. With the parabola \(x^2 = 6y\), the vertex is at the origin \((0,0)\).Here's how to proceed:

• Plot the vertex at the origin.
• Mark the focus at \((0, \frac{3}{2})\), just above the vertex.
• Draw the directrix, a horizontal line at \(y = -\frac{3}{2}\).

With these points established, sketch the parabola by drawing a symmetric curve that opens upwards towards the focus. It should mirror equally on either side of the y-axis, curving outwards as it moves from vertex towards the focus and beyond, approaching but never crossing, the directrix. This careful balance between elements makes your sketch precise and illustrative of the parabola's defining properties.