Problem 23 - Precalculus: Mathematics for Calculus

Question

An equation of a parabola is given. (a) Find the focus, directrix, and focal diameter of the parabola. (b) Sketch a graph of the parabola and its directrix. $$5 x+3 y^{2}=0$$

Step-by-Step Solution

Answer

Focus: \((-\frac{3}{20}, 0)\), Directrix: \(x = \frac{3}{20}\), Focal Diameter: \(\frac{3}{5}\).

Step 1: Rewrite the Equation in Standard Form

The given equation is \(5x + 3y^2 = 0\). Let's rewrite this in the form that resembles the standard form of a parabola \(y^2 = 4px\). To do this, divide every term by 5 to isolate \(x\) on one side:\[ x = -\frac{3}{5} y^2 \].

Step 2: Identify the Direction of the Parabola

Since the equation is in the form \(x = ay^2\) (\(a < 0\)), the parabola opens to the left. This is similar to \(y^2 = 4px\) with \(p\) being negative.

Step 3: Calculate the Focus of the Parabola

From \(y^2 = 4px\), we have \(4p = -\frac{3}{5}\). Solving for \(p\), we get \(p = -\frac{3}{20}\). The focus of the parabola for a vertex at the origin \((0,0)\) is \((p,0)\). Thus, the focus is \((-\frac{3}{20}, 0)\).

Step 4: Determine the Directrix of the Parabola

The directrix of a parabola \(y^2 = 4px\) is \(x = -p\). Since the vertex is at the origin and \(p = -\frac{3}{20}\), the directrix is \(x = \frac{3}{20}\).

Step 5: Calculate the Focal Diameter

The focal diameter is the length of the segment through the focus parallel to the directrix and equals \(|4p|\). Therefore, the focal diameter is \(\left| 4 \times -\frac{3}{20} \right| = \frac{3}{5}\).

Step 6: Sketch the Parabola and Directrix

Plot the vertex at the origin, focus at \((-\frac{3}{20},0)\), and the directrix \(x = \frac{3}{20}\). Draw the parabola opening left as a reflection around the line \(y = 0\). The focal width of \(\frac{3}{5}\) should be shown across the focus.

Key Concepts

Focus of a Parabola

The focus of a parabola is a point inside the curve that has a unique reflective property. Every point on the parabola is equidistant from the focus and a line known as the directrix. To find the focus of a parabola that opens sideways, like in the equation presented as \[ x = -\frac{3}{5} y^2 \],we derive from the standard form \( y^2 = 4px \).Here, \( 4p = -\frac{3}{5} \),which simplifies to \( p = -\frac{3}{20} \).The focus is then located at \((p, 0)\),which in this case is \((-\frac{3}{20}, 0)\).
A parabola's focus acts like a sort of 'mirror' where light or sound waves emanating from it reflect out parallel to the axis. This property is essential in the design of satellite dishes and microphones, helping to focus signals.

Directrix of a Parabola

The directrix of a parabola is a crucial line that helps define the shape and position of the parabola. It is found as the line that is equidistant from all points on the curve of the parabola. For the equation \( y^2 = 4px \),the directrix is the line \( x = -p \).
In the case of this parabola, because \( p = -\frac{3}{20} \),the directrix is given by the line \( x = \frac{3}{20} \).
The directrix serves as a balancing point against the focus. While the focus is a point, the directrix is an entire line, and together, they create the field where the parabola exists and defines its reflective properties, as noted previously.

Focal Diameter

The focal diameter, sometimes called the latus rectum, is a segment that passes through the focus and is perpendicular to the axis of symmetry of the parabola. Its length is \(|4p|\).In our equation,\( p = -\frac{3}{20} \),thus the focal diameter is \(\left| 4 \times -\frac{3}{20} \right| = \frac{3}{5} \).
This segment's length tells you that the width of the parabola at its focus is constant at \(\frac{3}{5} \).Understanding the focal diameter is essential as it expresses how 'wide' the parabola is at its most open point, and it is an intrinsic part of the parabolic structure. This concept is particularly useful when plotting graphs and can help in technological fields such as optics and engineering.

Graphing Parabolas

Graphing a parabola involves several steps that reflect the relationship between the focus, the directrix, and the direction of opening. To better understand this, consider our transformed parabola\( x = -\frac{3}{5} y^2 \).

**Graphing Steps:**

• First, plot the vertex of the parabola. For this equation, the vertex is at \((0, 0)\).
• Next, plot the focus which lies at \((-\frac{3}{20}, 0)\).
• Then, draw the directrix line \( x = \frac{3}{20} \),which runs parallel to the \(y\)-axis.
• Draw the parabola opening to the left, ensuring it curves symmetrically around the focus.
• Finally, sketch the parabola's focal diameter at \(\frac{3}{5} \)for visual representation.

Graphing parabolas effectively involves understanding these elements that define its shape and size. In visualizing the graph, these properties help in relating algebraic expressions to geometric realities, preparing students for practical applications.