Problem 14 - Algebra and Trigonometry

Question

Find the focus, directrix, and focal diameter of the parabola, and sketch its graph. $$x=\frac{1}{2} y^{2}$$

Step-by-Step Solution

Answer

Focus at (1, 0), directrix is x = -1, focal diameter is 2.

Step 1: Identify the Parabola Type

The given equation is \(x = \frac{1}{2}y^2\). This is of the form \(x = a y^2\), which indicates it is a horizontal parabola opening to the right or left, depending on the value of \(a\).

Step 2: Determine Key Parameters

To find the key parameters for the parabola, compare \(x = \frac{1}{2}y^2\) with the standard form \(x = a(y - k)^2 + h\) for horizontal parabolas. Here, \(a = \frac{1}{2}\), \(h = 0\), and \(k = 0\).

Step 3: Find the Focus

The focus of a horizontal parabola can be found using the formula: \( (h + \frac{1}{4a}, k) \). Since \(a = \frac{1}{2}\), \(h = 0\), and \(k = 0\), the focus is at \( (0 + \frac{1}{4 \times \frac{1}{2}}, 0) = (1, 0)\).

Step 4: Determine the Directrix

The directrix of a horizontal parabola can be given by the line \( x = h - \frac{1}{4a} \). Substitute \(h = 0\) and \(a = \frac{1}{2}\) to get the directrix: \(x = 0 - \frac{1}{4 \times \frac{1}{2}} = -1\).

Step 5: Calculate the Focal Diameter

The focal diameter, or the length of the latus rectum, is \(\frac{1}{|a|}\). Here, \(a = \frac{1}{2}\), so the focal diameter is \(\frac{1}{\frac{1}{2}} = 2\).

Step 6: Sketch the Graph

Draw the directrix line \(x = -1\). Plot the focus point at \((1, 0)\) and the vertex at \((0, 0)\). Sketch the parabola opening to the right, equidistant from the directrix and focus, with width defined by the focal diameter of 2.

Key Concepts

Focus of a Parabola

In a parabola, the focus is a crucial point that defines its shape and direction. Precisely, the focus lies inside the curve of the parabola.The intimate connection between the focus and the rest of the parabola is that the distance from any point on the parabola to the focus is equal to its distance to the directrix. For our example, the focus of the parabola from the equation \(x = \frac{1}{2}y^2\) is found at \( (1, 0) \). To find the focus of a parabola with the equation in the standard form \(x = a(y-k)^2 + h\):

• Use \(a = \frac{1}{2}\), \(h = 0\), and \(k = 0\) for our example.
• Compute the focus using \((h + \frac{1}{4a}, k)\).
• For this equation, the focus simplifies to \((1, 0)\).

The focus directly influences the direction in which the parabola opens, and for our example, it indicates that the parabola opens to the right.

Directrix of a Parabola

The directrix is a fascinating component of a parabola that directly influences its shape. Think of it as an invisible line helping define the boundaries and orientation of the parabola.In terms of placement, the directrix serves as the baseline from which the curve is reflected. It is perpendicular to the axis of symmetry of the parabola. When calculating the directrix for the parabola \(x = \frac{1}{2}y^2\), the directrix is found by using the formula: \( x = h - \frac{1}{4a} \). Here's the calculation for our example:

• Plug in \(h = 0\) and \(a = \frac{1}{2}\) into the formula.
• The directrix becomes \(x = 0 - \frac{1}{4 \times \frac{1}{2}} = -1\).

Thus, the directrix is a simple vertical line placed at \(x = -1\), ensuring each point on the parabola is equidistant to the focus \((1, 0)\) and to this line.

Focal Diameter of a Parabola

The focal diameter, often referred to as the length of the latus rectum, is a straightforward measurement crucial to understanding the breadth of a parabola. This dimension passes through the focus and is perpendicular to the axis of symmetry.For the parabola \(x = \frac{1}{2}y^2\), its focal diameter can be calculated using the formula:\( \frac{1}{|a|} \). Using our given \(a = \frac{1}{2}\), here's what you need to do:

• Take the reciprocal of the absolute value of \(a\).
• The result is \(\frac{1}{\frac{1}{2}} = 2\).

This means the focal diameter measures 2 units, allowing the parabola to expand the same distance in both directions from the focus. Understanding the focal diameter helps in sketching the parabola precisely, ensuring it captures the accurate width at its narrowest point.