Problem 26 - Algebra for College Students

Question

For Problems \(1-30\), find the vertex, focus, and directrix of the given parabola and sketch its graph. $$ x^{2}-4 x+4 y-4=0 $$

Step-by-Step Solution

Answer

Vertex: (2, 2), Focus: (2, 1), Directrix: y = 3.

Step 1: Rewrite in Standard Form

The given equation is \( x^{2} - 4x + 4y - 4 = 0 \). First, isolate the terms involving \( y \) to form an equation suitable for completing the square for \( x \). We get:\[ x^2 - 4x = -4y + 4 \]

Step 2: Complete the Square

Complete the square for the expression \( x^2 - 4x \). Add and subtract \( 4 \) inside the equation to balance it:\[ (x^2 - 4x + 4) - 4 = -4y + 4 \]This results in:\[ (x - 2)^2 = -4y + 8 \]

Step 3: Solve for y

Bring all terms with \( y \) to one side to reveal the standard form of a vertical parabola:\[ (x - 2)^2 = -4(y - 2) \]

Step 4: Identify the Parabola Components

The equation is now in the form \( (x - h)^2 = 4p(y - k) \), with \( h = 2 \), \( k = 2 \), and \( 4p = -4 \). Thus, \( p = -1 \). The vertex is \((h, k) = (2, 2)\).

Step 5: Find the Focus

The focus\((h, k + p)\) is \((2, 2 + (-1))\), which is at \((2, 1)\).

Step 6: Find the Directrix

The directrix is \( y = k - p \), which translates to \( y = 2 - (-1) = 3 \).

Step 7: Sketch the Graph

Draw the vertex at (2, 2) and plot the focus at (2, 1). Draw the directrix as a horizontal line at \( y = 3 \). Sketch the parabola opening downwards, equidistant from the focus and directrix.

Key Concepts

Vertex of a Parabola

The vertex of a parabola is the point where the curve changes direction. It acts as the "turning" point and can be thought of as the point of symmetry for the parabola. For the equation in standard form \((x-h)^2 = 4p(y-k)\), the vertex is given by the coordinates \((h, k)\). In this form, the parabola is oriented vertically, and the point \((h, k)\) lies on the axis of symmetry.

• In our example, the parabola \((x-2)^2 = -4(y-2)\) has the vertex at \((h, k) = (2, 2)\).
• The vertex tells us that the parabola opens downwards if \(4p\) is negative, as it does here with \(4p = -4\).
• To find the vertex from a general equation like \(ax^2 + bx + c = y\), one useful trick is to complete the square.

In general, finding the vertex allows us to quickly sketch and understand the parabola's orientation and position in the coordinate plane. Knowing the direction the parabola opens (up or down) is crucial when visualizing its graph.

Focus of a Parabola

The focus of a parabola is a critical point that leads to its definition. A parabola is the set of all points equidistant from a fixed point, known as the focus, and a line known as the directrix. The focus lies inside the parabola.

• In the equation \((x-h)^2 = 4p(y-k)\), the focus is located at \((h, k + p)\) for a vertically oriented parabola.
• From our example equation, \((x-2)^2 = -4(y-2)\), the focus is at \((2, 1)\) because \(p = -1\).
• The focus's distance from the vertex is determined by \(p\), indicating the width and shape of the parabola's "bowl."

Understanding the focus is crucial as it helps to comprehend how light or sound waves reflect off the parabolic shape. For instance, satellite dishes and car headlights use parabolas to focus signals or light onto a single point.

Directrix of a Parabola

The directrix of a parabola is a line that, along with the focus, helps in defining the parabola itself. This line is symmetrical about the axis of symmetry of the parabola and lies on the opposite side of the curve from the focus.

• For a vertically oriented parabola in the form \((x-h)^2 = 4p(y-k)\), the directrix is the horizontal line \(y = k - p\).
• In the example we've solved, with equation \((x-2)^2 = -4(y-2)\), the directrix is located at \(y = 3\).
• This line serves as a guideline to ensure that any point on the parabola is equidistant from the focus and the directrix.

The directrix complements the focus, showing the parabola's symmetrical properties and helping in accurately sketching its graph. Understanding the directrix renders a full picture of how the parabola behaves and where it lies in the coordinate system.