Problem 46 - Precalculus with Limits

Question

In Exercises 33-46, find the vertex, focus, and directrix of the parabola, and sketch its graph. \(y^2 - 4y - 4x = 0\)

Step-by-Step Solution

Answer

The vertex of the parabola is (-1, 2), the focus is (0, 2) and the directrix is x = -2. The filled sketch should show a parabola opening to the right, with the vertex at (-1,2), the focus inside the parabola and the directrix as a line on its left.

Step 1: Convert the equation to vertex form

We start by rewriting the given equation \(y^2 - 4y - 4x = 0 \) into a standard form that hints about the vertex of the parabola. Since the square term is for \(y\), the parabola opens either to the left or right. We complete squares for \(y\) to make it into standard form. \((y - 2)^2 = 4x + 4 \) which simplifies to \((y - 2)^2 = 4(x + 1)\). Which tells us that the recipe for a parabola is \( (y-k)^2 = 4p(x-h) \).

Step 2: Identify the Vertex

By matching the equation \((y - 2)^2 = 4(x + 1)\) with the standard equation \((y-k)^2 = 4p(x-h)\), we identify the vertex of the parabola as \(h = -1\) and \(k = 2\). So, the vertex is \((-1, 2)\).

Step 3: Find the Value of p

Next, identify the value of p which affects the direction the parabola opens and controls the width of the parabola. From our equation \((y - 2)^2 = 4(x + 1)\), p is equal to 1.

Step 4: Identify the Focus and Directrix

In the equation \((y - k)^2 = 4p(x - h)\), the focus is \((h+p, k)\) and the directrix is \(x = h - p\). With \(h = -1, k = 2\) and \(p = 1\), we can deduce the focus as \((-1 +1, 2) = (0, 2)\) and the directrix as \(x = -1 - 1 = -2\).

Step 5: Sketch the graph

With these values, we can now draw the graph of the parabola. Place the vertex at point \((-1,2)\), draw the directrix line at \(x = -2\) and mark the focus point at \(0,2\). Draw a parabola opening to the right, with the focus on the interior and the directrix on the exterior.

Key Concepts

Vertex

The vertex of a parabola is a significant point, as it represents the location where the parabola either reaches its maximum or minimum value depending on the direction in which it opens. In the given equation \[(y - 2)^2 = 4(x + 1),\]the vertex is deduced by comparing it with the standard form of a parabola's equation, which is \[(y-k)^2 = 4p(x-h).\]Here, the vertex \((h, k)\)is identified by locating \(h = -1\)and \(k = 2\).Therefore, the vertex of the parabola is at \((-1, 2)\).This point is crucial since it serves as the parabola's pivot point, influencing its shape and direction.

• The coordinates \((-1, 2)\) inform us where to start drawing the parabola.
• Understanding the vertex allows us to establish the symmetry of the parabola.

It’s essential for finding other characteristics like focus and directrix.

Focus

The focus of a parabola is another vital aspect as it contributes to the unique reflective property of the curve. All points of the parabola are equidistant from this point and the directrix. From the standard form equation \((y-k)^2 = 4p(x-h),\) we ascertain the focus using the formula \((h + p, k)\).Given \(p = 1\), \(h = -1\), and \(k = 2\), we determine that the focus of the parabola \((h+p, k)\)is \((0, 2)\).

• Positioning \((0, 2)\) inside the parabola ensures that the curve bends outward away from the directrix, focusing towards this point.
• The focus lies on the interior of the parabola, and every line drawn from any point on the parabola to this focus reflects evenly across the axis of symmetry, heading back to the directrix.

Directrix

The directrix is an invisible line that assists in defining the parabola's shape and focus. It is crucial as it acts together with the focus to maintain the set distances that define the curve. For a parabola represented by \((y-k)^2 = 4p(x-h),\) the directrix can be determined using the formula \(x = h - p\).

Directrix for our Parabola

Given \(h = -1\) and \(p = 1\), we have that the directrix is at:\[x = -1 - 1 = -2.\]This means the directrix is a vertical line along \(x = -2\).

• This line does not touch the parabola but remains equidistant from the focus.
• It helps determine how sharply the parabola opens.

The directrix plays an indispensable role in defining the parabola’s geometric set of curves as all points are equidistant between it and the focus.