Problem 21 - Algebra and Trigonometry

Question

\(11-22\) . Find the focus, directrix, and focal diameter of the parabola, and sketch its graph. $$ 5 x+3 y^{2}=0 $$

Step-by-Step Solution

Answer

The focus is \((-\frac{5}{12}, 0)\), the directrix is \(x = \frac{5}{12}\), and the focal diameter is \(\frac{5}{3}\).

Step 1: Rearrange the Equation

The given equation is \(5x + 3y^2 = 0\). Start by isolating \(x\) in terms of \(y\):\[5x = -3y^2\]Divide both sides by 5:\[x = -\frac{3}{5}y^2\]

Step 2: Identify the Form of the Parabola

The equation \(x = -\frac{3}{5}y^2\) is similar to the standard form of a parabola \(x = ay^2\).Here, \(a = -\frac{3}{5}\). Since \(a\) is negative, the parabola opens to the left.

Step 3: Calculate the Focus and Directrix

For a parabola in the form \(x = ay^2\), the focus is at \(\left(\frac{1}{4a}, 0\right)\), and the directrix is the vertical line \(x = -\frac{1}{4a}\).Substituting \(a = -\frac{3}{5}\):\[\frac{1}{4a} = \frac{1}{4 \times -\frac{3}{5}} = -\frac{5}{12}\]- The focus is at \((-\frac{5}{12}, 0)\).- The directrix is the line \(x = \frac{5}{12}\).

Step 4: Determine the Focal Diameter

The focal diameter of the parabola is \(\frac{|1|}{|a|}\).\[\frac{1}{\left| -\frac{3}{5} \right|} = \frac{5}{3}\]Thus, the focal diameter is \(\frac{5}{3}\).

Step 5: Sketch the Graph

To sketch the parabola:1. The vertex is at the origin (0,0).2. The parabola opens to the left since \(a\) is negative.3. Plot the focus at \((-\frac{5}{12}, 0)\).4. Draw the directrix as the vertical line \(x = \frac{5}{12}\).5. The parabola will be symmetric about the y-axis, passing through the vertex and approaching the directrix without intersecting it.

Key Concepts

Focus of a Parabola

The focus of a parabola is a special point that has a unique geometric significance. This point is located on the interior of the parabola, and every point on the parabola is equidistant to the focus and a line called the directrix.
For the equation we are looking at, the focus can be found using the formula for a parabola \(x = ay^2\), where the location of the focus is at \((\frac{1}{4a}, 0)\).
In our example, \(a = -\frac{3}{5}\), so:

• Substituting the value of \(a\) into the formula gives the focus located at \((-\frac{5}{12}, 0)\).

This means that when the graph is plotted, the focus will be to the left of the origin, following the direction the parabola opens.

Directrix of a Parabola

The directrix of a parabola is an essential line for defining its shape. It is always perpendicular to the axis of symmetry of the parabola and located on the opposite side from the focus.
For the standard form \(x = ay^2\), the directrix line is determined by \(x = -\frac{1}{4a}\).

• Inserting \(a = -\frac{3}{5}\) produces the equation for the directrix: \(x = \frac{5}{12}\).

As we sketch the parabola, this vertical line plays a key role in maintaining the symmetry of the shape, acting as a boundary that the parabola approaches but never crosses. Knowing the directrix assists in getting a precise plot of how the parabola sits relative to the y-axis.

Focal Diameter

The focal diameter, sometimes referred to as the latus rectum, is another fascinating characteristic of parabolas. This line segment passes through the focus, parallel to the directrix, and its endpoints lie on the parabola itself.
For parabolas expressed as \(x = ay^2\), the length of the focal diameter is given by \(\frac{1}{|a|}\).

• In this example, using \(a = -\frac{3}{5}\) produces a focal diameter of \(\frac{5}{3}\).

The focal diameter measures the width of the parabola at the focus, providing an understanding of how 'wide' the parabola is at its center. This aids in sketching the parabola accurately and highlights the parabola's symmetry from side to side.

Graph of a Parabola

Graphing a parabola brings all its properties together in a visual form, helping to understand its structure and orientation.
To graph the example \(5x + 3y^2 = 0\), follow these steps:

• The vertex, often the starting point, is at the origin \((0,0)\).
• Given \(a = -\frac{3}{5}\), the parabola opens to the left.
• Mark the focus at \((-\frac{5}{12}, 0)\) on the graph.
• Draw the directrix line at \(x = \frac{5}{12}\).
• Sketch the parabola to reflect symmetry about the vertex, ensuring no crossing over to the directrix.

When graphing, note that the parabola's openness tells us about its orientation (leftwards in this case), and features like the vertex, focus, and directrix guide its curvature correctly. Visualizing these elements allows for an intuitive grasp of the parabola's geometry.