Problem 12 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

Focus: \((0, -\frac{1}{8})\); Directrix: \(y = \frac{1}{8}\); Focal diameter: \(\frac{1}{2}\).

Step 1: Rewrite the Equation in Standard Form

The standard form of a parabola that opens vertically is \( y = ax^2 + bx + c \). In this equation, we have \( y = -2x^2 \), which is already in standard form with \( a = -2 \), \( b = 0 \), and \( c = 0 \).

Step 2: Identify Parameters

For the parabola \( y = ax^2 \), the focus is at \( (0, \frac{1}{4a}) \), and the directrix is \( y = -\frac{1}{4a} \). Here, \( a = -2 \), so \( \frac{1}{4a} = \frac{1}{-8} = -\frac{1}{8} \).

Step 3: Find the Focus

Using the value \( \frac{1}{4a} \), the focus of the parabola is at \( (0, -\frac{1}{8}) \).

Step 4: Determine the Directrix

The directrix of the parabola is the line \( y = -\frac{1}{4a} \), which evaluates to \( y = \frac{1}{8} \).

Step 5: Calculate the Focal Diameter

The focal diameter is the absolute value of \( \frac{1}{a} \). With \( a = -2 \), \( \frac{1}{a} = \frac{1}{-2} = -\frac{1}{2} \), so the focal diameter is \( \frac{1}{2} \).

Step 6: Sketch the Graph

The parabola opens downward since \( a = -2 \). The vertex is at the origin \( (0, 0) \), the focus is at \( (0, -\frac{1}{8}) \), and the directrix is the line \( y = \frac{1}{8} \). The graph is a downward-opening parabola with the vertex at the origin.

Key Concepts

Focus of a Parabola

In the world of parabolas, the focus plays a crucial role in defining its shape and orientation. The focus of a parabola is a specific point located inside the curve. This point is so special because every point on the parabola is equidistant from the focus and a line called the directrix (which we'll discuss in the next section).
For the parabola defined by the equation \( y = -2x^2 \), finding the focus starts by identifying the standard form parameter \( a \) from the equation. For our parabola, \( a = -2 \). A helpful formula for finding the focus in a vertically oriented parabola (one that opens up or down) is \( (0, \frac{1}{4a}) \).
Using this formula, the focus for \( y = -2x^2 \) is calculated as follows:

• Identify \( a = -2 \)
• Calculate \( \frac{1}{4a} = \frac{1}{-8} = -\frac{1}{8} \)
• Focus is at \( (0, -\frac{1}{8}) \)

This tells us that our parabola reaches its deepest point at \( (0, -\frac{1}{8}) \), directing downward toward its focus.

Directrix of a Parabola

Equally important to the focus, the directrix serves as a guiding line for the parabola. While the focus is a point inside the curve, the directrix is a straight line outside of it. All points from the parabola maintain equal distance between themselves, the focus, and this specific line.
For a parabola opening up or down, the directrix' equation is \( y = -\frac{1}{4a} \). In our case where \( a = -2 \), substituting into the formula provides:

• Directrix Calculation: \( y = \frac{1}{8} \)

This result means the line \( y = \frac{1}{8} \) acts as the directrix. It's situated above the vertex at the origin for the parabola \( y = -2x^2 \), enforcing the downward-opening nature of the curve by balancing the internal focus.

Focal Diameter

The focal diameter of a parabola measures the width of the parabola at its focus, telling us how 'open' or 'narrow' the parabola's curve appears. It represents the distance between two points on the parabola that are collinear with the focus.
To find the focal diameter, we use the formula \( \frac{1}{a} \). With \( a = -2 \) in our given parabola equation, we have:

• \( \frac{1}{a} = \frac{1}{-2} = -\frac{1}{2} \)
• The absolute value gives us a focal diameter of \( \frac{1}{2} \)

This value reflects how broad the parabola opens around its focus. The smaller the focal diameter, the sharper or tighter the curve is near the focus, as seen in this specific parabola example.