Problem 12 - Precalculus with Calculus Previews

Question

Find the vertex, focus, directrix, and axis of the given parabola. Graph the parabola. \((x-2)^{2}+y=0\)

Step-by-Step Solution

Answer

Vertex: (2, 0). Focus: (2, -1/4). Directrix: y = 1/4. Axis: x = 2.

Step 1: Write the equation in standard form

The given equation is \((x-2)^2 + y = 0\). To put it in standard form, \((x-h)^2 = 4p(y-k)\), we rearrange it to solve for \(y\): \(y = -(x-2)^2\). This can be rewritten as \((x-2)^2 = -1(y-0)\). In this form, we identify: \(h = 2\), \(k = 0\), and \(4p = -1\).

Step 2: Identify the vertex

In the standard form \((x-h)^2 = 4p(y-k)\), the vertex of the parabola is located at \((h, k)\). From Step 1, since \(h = 2\) and \(k = 0\), the vertex is \((2, 0)\).

Step 3: Calculate the value of p

We found that \(4p = -1\) from the standard form. Solving for \(p\), we get \(p = -\frac{1}{4}\). This value determines the orientation and position of the focus and directrix.

Step 4: Determine the focus

The focus of a parabola \((x-h)^2 = 4p(y-k)\) is located at \((h, k + p)\). Substituting the known values, we find the focus is at \((2, 0 + (-\frac{1}{4})) = (2, -\frac{1}{4})\).

Step 5: Find the equation of the directrix

The directrix is a line located at \(y = k - p\). Using the values from previous steps, the equation becomes \(y = 0 - (-\frac{1}{4}) = \frac{1}{4}\).

Step 6: Determine the axis of symmetry

The axis of symmetry of a parabola \((x-h)^2 = 4p(y-k)\) is the vertical line \(x = h\). Based on our equation, it's \(x = 2\). This line passes through the vertex and is equidistant from the focus and directrix.

Step 7: Sketch the graph

Draw a vertical line for the axis of symmetry at \(x = 2\), plot the vertex at \((2, 0)\), and the focus at \((2, -\frac{1}{4})\). The directrix is a horizontal line at \(y = \frac{1}{4}\). Since this is a downward opening parabola (as \(p < 0\)), sketch the parabolic curve opening downwards starting from the vertex.

Key Concepts

Vertex

The vertex of a parabola is a crucial point where the curve changes direction. For a parabola given in the standard form

• \((x-h)^2 = 4p(y-k)\)

the vertex is located at

• \((h, k)\).

In our example, starting with the equation \((x-2)^2 + y = 0\), we put the equation in standard form and identified

• \(h = 2\)
• \(k = 0\)

Thus, the vertex is located at \((2, 0)\).

This point is the highest (or lowest) point when a parabola is opening downwards (or upwards). Since our parabola opens downward, the vertex at \((2, 0)\) is the highest point on the curve. Understanding the vertex allows us to know where this parabola peaks or bottoms out.

Focus

The focus of a parabola is always located inside the curve. It is one of the points used to define the parabola itself. The distance between the vertex and the focus is given by

• \(p\)

from the standard form

• \((x-h)^2 = 4p(y-k)\).

In our specific example, we have calculated

• \(p = -\frac{1}{4}\)

which tells us that the focus is \(\frac{1}{4}\) unit below the vertex because \(p\) is negative, indicating a downward opening parabola. This places our focus at

• \((2, -\frac{1}{4})\).

The focus is important because any ray entering parallel to the axis of symmetry will converge at this point. This unique property is often used in optics and satellite dishes.

Directrix

The directrix of a parabola is a line that, together with the focus, helps define a parabola. This line is located opposite the direction of the opening from the vertex. The relationship for the directrix in a standard form parabola

• \((x-h)^2 = 4p(y-k)\)

is given by

• \(y = k - p\).

Based on the given problem, with

• \(k = 0\)
• \(p = -\frac{1}{4}\)

the equation of the directrix becomes

• \(y = 0 - (-\frac{1}{4}) = \frac{1}{4}\).

This horizontal line runs parallel above the vertex and represents the balance point for the parabola. Its value is used to keep the focus equivalent on the other side. Note how equidistant every point of the parabola is to the focus and directrix.

Axis of symmetry

The axis of symmetry is an imaginary line that runs vertically through the vertex of the parabola. This line divides the parabola into two satisfactory mirror images. In the standard parabola form

• \((x-h)^2 = 4p(y-k)\)

the axis of symmetry is given as

• \(x = h\).

In the context of our problem, with

• \(h = 2\)

we see the axis of symmetry is the vertical line

• \(x = 2\).

This crucial line ensures that the parabola opens symmetrically around the axis. It acts like a balance line, dividing the shape equally, and aligning the focus and directrix directly opposite to each other perpendicularly. Understanding symmetry helps predict the parabola's form and guides in sketching its curve. The concept might even find application in designs and aesthetic tasks where balance is key.