Problem 18 - Precalculus with Calculus Previews

Question

Find the vertex, focus, directrix, and axis of the given parabola. Graph the parabola. \(y^{2}-4 y-4 x+3=0\)

Step-by-Step Solution

Answer

Vertex: \\(-\frac{1}{4}, 2\\), Focus: \\(\frac{3}{4}, 2\\), Directrix: \\(-\frac{5}{4}\\), Axis: \\(y = 2\\).

Step 1: Rewrite the Equation in Standard Form

The given equation is: \[ y^2 - 4y - 4x + 3 = 0 \]To rewrite it in standard form, first isolate the terms involving "x":\[ y^2 - 4y + 3 = 4x \]This resembles the conic form \((y - k)^2 = 4p(x - h)\). Now, we complete the square for the "y" terms.

Step 2: Complete the Square

Take the terms \(y^2 - 4y\) and complete the square:- Identify the coefficient of "y", which is -4.- Half of -4 is -2, and its square is 4.Add and subtract 4 inside the equation to complete the square:\[ y^2 - 4y + 4 - 1 = 4x \]Thus, we rewrite it as:\[ (y - 2)^2 = 4x + 1 \]Or rearrange it to:\[(y - 2)^2 = 4(x + \frac{1}{4}) \]This is now in the standard form \((y-k)^2 = 4p(x-h)\).

Step 3: Identify the Vertex

In the standard form \((y-k)^2 = 4p(x-h)\), the vertex is \((h, k)\). From our equation, \( (y-2)^2 = 4(x + \frac{1}{4}) \), so the vertex is at:\[\left(-\frac{1}{4}, 2\right)\]

Step 4: Find the Focus

The parameter "p" defines the distance from the vertex to the focus. We know \(4p = 4\), so \(p = 1\).Since the parabola opens to the right (positive x-direction), the focus from the vertex \((-\frac{1}{4}, 2)\) is:\[\left(-\frac{1}{4} + 1, 2\right) = \left(\frac{3}{4}, 2\right)\]

Step 5: Determine the Directrix

The directrix is a vertical line opposite the focus from the vertex, at distance "p" from the vertex.For this parabola, the directrix is:\[ x = -\frac{1}{4} - 1 = -\frac{5}{4}\]

Step 6: Identify the Axis

The axis of symmetry is the line that passes through the vertex and focus, which is parallel to the "y-axis" given that it's a horizontal parabola.Thus, the axis is:\[ y = 2 \]

Step 7: Graph the Parabola

To graph the parabola:- Plot the vertex at \((-\frac{1}{4}, 2)\).- Draw the focus at \((\frac{3}{4}, 2)\).- Sketch the parabola opening to the right, symmetrically around the line \(y = 2\).- Draw the vertical directrix at \(x = -\frac{5}{4}\).

Key Concepts

Vertex of a Parabola

The vertex of a parabola is a crucial point that acts as the "turning point" of the curve. Think of it as the point where the parabola changes direction. For parabolas in the form

• y-k)^2 = 4p(x-h)

the vertex is the point

• (h, k)

. In our standard form equation (y-2)^2 = 4(x + \(\frac{1}{4}\)), the vertex is located at (-\(\frac{1}{4}\), 2). Here, the vertex ties together with other elements of the parabola, acting as a bridge to finding the focus and directrix.

Focus and Directrix

The focus and directrix have unique roles in defining a parabola. The focus is a point, while the directrix is a line, and they together define all the points that form the parabola. Every point on the parabola is equidistant from the focus and the directrix line.
For our equation, once we isolate in the form

• (y-k)^2 = 4p(x-h)

and find p, we locate these elements.Given that 4p = 4, so p = 1:

• The focus: (\(\frac{3}{4}\), 2)
• The directrix: x = -\(\frac{5}{4}\)

These coordinates mean that if you draw a perpendicular line from any point on the parabola to the directrix or to the focus, both lengths should be the same.

Completing the Square

Completing the square is a handy technique to transform a quadratic equation into a standard form that is easier to understand and use. The goal of this method is to create a perfect square trinomial on one side of the equation, making it easier to identify the vertex.In our equation

• y^2 - 4y - 4x + 3 = 0

, you isolate the y-terms and find the completing square terms:

• Identify the coefficient of y, which is -4
• Half it and square it to keep balance
• Rewrite: (y - 2)^2 = 4x + 1

Ultimately, this step allowed us to clearly see the vertex is (-\(\frac{1}{4}\), 2) and continued onto mapping out focus and directrix.

Graphing Parabolas

Graphing a parabola helps visually understand its properties from the standard form equation. Once the characteristics like vertex, focus, and directrix are found, sketching becomes more straightforward.Start by plotting the

• Vertex at (-\(\frac{1}{4}\), 2)
• Focus at (\(\frac{3}{4}\), 2)
• Directrix line x = -\(\frac{5}{4}\)

Next, use symmetry to guide the parabola around the axis parallel to the y-axis at y = 2. The direction the parabola opens is determined by the locus of points equidistant from the focus to the directrix; in this case to the right.