Problem 15 - Precalculus: Mathematical for Calculus

Question

Find the focus, directrix, and focal diameter of the parabola, and sketch its graph. $$x^{2}+6 y=0$$

Step-by-Step Solution

Answer

Focus: \((0, -\frac{3}{2})\); Directrix: \(y = \frac{3}{2}\); Focal Diameter: 6.

Step 1: Rearrange the Equation

The given equation is \(x^{2} + 6y = 0\). We rearrange it to express \(y\) in terms of \(x\). Move \(6y\) to the right side: \[x^2 = -6y\]. Now, divide each side by -6 to solve for \(y\):\[y = -\frac{x^2}{6}.\]

Step 2: Identify the Standard Form

The equation \(y = -\frac{x^2}{6}\) is of the form \(y = -\frac{1}{4p}x^2\). Here, comparing with \(y = -\frac{1}{4p}x^2\), we find \(\frac{1}{4p} = \frac{1}{6}\), which implies \(4p = 6\).

Step 3: Solve for \(p\)

From the equation \(4p = 6\), solve for \(p\):\[p = \frac{6}{4} = \frac{3}{2}.\] This \(p\) value will be used to determine the focus and the directrix of the parabola.

Step 4: Determine the Focus and Directrix

For the parabola \(y = -\frac{1}{4p}x^2\), the focus is at \((0, -p)\) and the directrix is \(y = p\). Substituting \(p = \frac{3}{2}\), the focus is at \((0, -\frac{3}{2})\), and the directrix is \(y = \frac{3}{2}\).

Step 5: Find the Focal Diameter

The focal diameter is \(|4p|\). Since we determined \(p = \frac{3}{2}\), the focal diameter is \(|4(\frac{3}{2})| = |6| = 6\).

Step 6: Sketch the Parabola

To sketch the parabola, plot the focus \((0, -\frac{3}{2})\) and the directrix \(y = \frac{3}{2}\). The parabola opens downwards because the coefficient of \(x^2\) in \(y = -\frac{x^2}{6}\) is negative. Draw the parabola with vertex at the origin \((0, 0)\), extending downward symmetrically around the \(y\)-axis.

Key Concepts

Focus

In a parabola, the focus is a special point that defines the curve's shape and direction. It is one of two critical components—the other being the directrix—that determine what the parabola looks like and how it is oriented. For any parabola described by the equation

• \(y = \frac{1}{4p}x^2\) (when it opens upward or downward)
• \(x = \frac{1}{4p}y^2\) (when it opens to the left or right)

its focus will have specific coordinates relative to its vertex. In a parabola that opens vertically, such as our example \(y = -\frac{x^2}{6}\), the focus lies directly along the axis of symmetry—either above or below the vertex depending on the direction the parabola opens.

To find the focus for the downward-opening parabola in our exercise:

• Determine \(p\), the distance from the vertex to the focus, from the equation \(y = -\frac{1}{4p}x^2\).
• Once \(p\) is calculated, the focus is at \((0, -p)\), in the case of our vertically oriented parabola.
• In this case, \(p\) is \(\frac{3}{2}\), so the focus is at \((0, -\frac{3}{2})\).

This point is where all paths reflected off the surface of the parabola converge, emphasizing its importance in applications like satellite dishes and flashlights.

Directrix

A directrix in a parabola serves as a guideline that, along with the focus, gives the parabola its unique set of properties. The directrix is always a line, and it is as significant as the focus in determining the shape of the parabola. For a vertically oriented parabola like the one we have in \(y = -\frac{x^2}{6}\), the directrix is a horizontal line located parallel to the \(x\)-axis.

In the equation form \(y = \frac{1}{4p}x^2\), the directrix is located at \(y = -p\) when the parabola opens upwards and \(y = p\) when it opens downwards.

• For our example, where \(p\) is \(\frac{3}{2}\), the equation for the directrix is \(y = \frac{3}{2}\).
• This means the horizontal line runs parallel to the \(x\)-axis, passing through the point \(y = \frac{3}{2}\).

The directrix helps form the geometric property that defines parabolas: each point on the parabola is equidistant from the focus and this fixed directrix line. This unique location and function of the directrix make it fundamental to understanding how parabolas behave and are used in various mathematical applications.

Focal Diameter

The focal diameter is another essential aspect of understanding the properties of a parabola. It concerns the width of the parabola at the level of the focus, providing insight into how "wide" or "narrow" the parabola is. Mathematically, the focal diameter is simply the absolute value of \(|4p|\).

What this means is:

• For any parabola given by \(y = \frac{1}{4p}x^2\), the focal diameter is the absolute distance across the parabola, passing through the focus.
• It ensures that the section of the parabola around the focus maintains symmetry.

For our parabola

• With \(p = \frac{3}{2}\), the focal diameter is computed as \(|4(\frac{3}{2})| = 6\).
• This width directly contributes to the curve's appearance, influencing whether it seems wide open or tightly closed.

The focal diameter is important in contexts such as optics and physics because it can affect how a parabola focuses or reflects light and sound waves.