Problem 11 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-12: Find the vertex, focus, and directrix of the parabola. Sketch its graph, showing the focus and the directrix. $$ x^{2}+20 y=10 $$

Step-by-Step Solution

Answer

The vertex is \((0, \frac{1}{2})\), focus is \((0, -\frac{9}{2})\), and directrix is \( y = \frac{11}{2} \).

Step 1: Rewrite the Equation

Start by rewriting the equation of the parabola so that it resembles the standard form. The given equation is \( x^2 + 20y = 10 \). Isolate the term with \( y \) on one side: \( 20y = -x^2 + 10 \). Then, divide by 20 to solve for \( y \): \( y = -\frac{1}{20}x^2 + \frac{1}{2} \).

Step 2: Identify the Vertex

The standard form of a parabola that opens vertically is \( y = ax^2 + bx + c \). In this equation, \( y = -\frac{1}{20}x^2 + \frac{1}{2} \), the vertex is at the point \((h, k)\), where \( h \) is 0 (since there is no \( x \) term) and \( k \) is the constant term, which is \( \frac{1}{2} \). Thus, the vertex is at \( (0, \frac{1}{2}) \).

Step 3: Find the Focus and Directrix

For a vertical parabola \( y = a(x-h)^2 + k \), the focus is at \( (h, k + \frac{1}{4a}) \). Here, \( a = -\frac{1}{20} \), so the focus is at \( (0, \frac{1}{2} + \frac{1}{4 \cdot (-1/20)}) = (0, \frac{1}{2} + (-5)) = (0, -\frac{9}{2}) \). The directrix is the line \( y = k - \frac{1}{4a} \), which is \( y = \frac{1}{2} + 5 = \frac{11}{2} \).

Step 4: Sketch the Graph

To sketch the graph, mark the vertex at \((0, \frac{1}{2})\). The parabola opens downward, as indicated by the negative \( a \) value. Plot the focus at \((0, -\frac{9}{2})\) and the directrix as the horizontal line \( y = \frac{11}{2} \). Draw the parabola passing through the vertex, opening downward, ensuring it is equidistant from the focus and directrix.

Key Concepts

Vertex

The vertex of a parabola is a key point that represents the location where the curve changes direction. It's a turning point and can be found easily once the equation is in standard form.

In our current example, the equation is transformed into the form: \[ y = -\frac{1}{20}x^2 + \frac{1}{2} \] This indicates the parabola opens vertically. In this format, the vertex is given by the pair \((h, k)\), where \(h\) comes from the \(x\) term, and \(k\) is the constant term.

Since there is no \(x\) term, \(h = 0\), and \(k = \frac{1}{2}\). Thus, the vertex is at \((0, \frac{1}{2})\). This point is crucial as it helps in plotting the parabola accurately.

Focus of a Parabola

The focus of a parabola is an important feature. It's one of the special points that determines the shape and orientation of the parabola.

Essentially, the focus is a point inside the parabola used along with the directrix. Every point on the parabola is equidistant from the focus and the directrix. For our case, we use the form:\[ y = a(x-h)^2 + k \] to locate the focus.

• The formula to determine the focus is \((h, k + \frac{1}{4a})\).
• Here, \(a = -\frac{1}{20}\), so \(k + \frac{1}{4a} = \frac{1}{2} + (-5) = -\frac{9}{2}\).
• Therefore, the focus is at \((0, -\frac{9}{2})\).

The focus helps us understand where the parabola is "aiming," so to speak.

Directrix of a Parabola

The directrix is a line, and it complements the focus. This line is perpendicular to the axis of symmetry of the parabola.

In essence, the parabola is defined such that any point on it is equidistant from the point called the focus and a line called the directrix.

For our particular parabola, with the form provided:

• We calculate the directrix using the formula \(y = k - \frac{1}{4a}\).
• Substituting the known values gives: \(y = \frac{1}{2} + 5 = \frac{11}{2}\).
• The line \(y = \frac{11}{2}\) is the directrix.

The directrix plays a crucial role during the sketching of the graph, as it, along with the focus, guides the shape of the parabola.

Graphing Parabolas

Graphing a parabola can initially seem daunting, but with a clear understanding of its components, it becomes straightforward. Here's a breakdown of how to approach it for this specific example:

Once the vertex, focus, and directrix are identified, plotting them provides a foundation for the graph. Following these steps will help:

• First, on a coordinate plane, plot the vertex \((0, \frac{1}{2})\).
• Next, mark the focus at \((0, -\frac{9}{2})\).
• Draw the directrix as the horizontal line \(y = \frac{11}{2}\).
• From the vertex, draw the curve of the parabola opening downwards, as dictated by the negative \(a\) value.

Ensuring the curve remains equidistant from the focus and directrix helps maintain its accuracy. By following these steps, you'll have a clear graph that visually represents the equation.