Problem 38 - Precalculus

Question

Find the vertex, focus, and directrix of the parabola. Then sketch the parabola. $$\left(x+\frac{1}{2}\right)^{2}=4(y-1)$$

Step-by-Step Solution

Answer

The vertex of the parabola is at (-1/2, 1), focus is at (-1/2, 2), and directrix is the line y = 0. It opens upwards.

Step 1: Identify the standard form

The given equation is in the standard form of a parabola that opens upwards or downwards, \( (x-h)^2 = 4p(y-k) \), where (h,k) is the vertex of the parabola. p is the distance from the vertex to the focus and also from vertex to the directrix. The sign of p tells us if the parabola opens upwards (positive) or downwards (negative).

Step 2: Find the vertex

By comparing \(\left(x+\frac{1}{2}\right)^{2}=4(y-1)\) with the standard form, we can see that h = -1/2 and k = 1, so the vertex of the parabola is (-1/2, 1).

Step 3: Find p

The coefficient of the y term is 4p. In our case, this is 4. Thus, 4p = 4 and solving this we get p = 1.

Step 4: Find the focus

For parabolas opening up or down, the focus is located p units above or below the vertex. The parabola opens upwards (because p is positive), so the focus is 1 unit above the vertex. This means that the focus is at (-1/2, 2).

Step 5: Find the directrix

The directrix is located p units below the vertex for parabolas that open up. This means the directrix is the line y = k - p = 1 - 1 = 0.

Step 6: Sketch the parabola

Now with all the key features - vertex (-1/2, 1), focus (-1/2, 2), and directrix y = 0, plot these on a graph. The parabola envelops the focus and is guided away from the directrix. It should be symmetric about the line through the focus and perpendicular to the directrix (the axis of symmetry).

Key Concepts

Vertex

In the context of a quadratic expression or parabola, the vertex refers to the point where the curve turns or where it is at its peak or lowest point, depending on its orientation. A parabola has a U-shaped curve. This point is crucial because it represents the highest or lowest point on the graph of a quadratic function.
The vertex for a parabola given by the equation \((x-h)^2 = 4p(y-k)\) is at the point (h, k). Here, 'h' and 'k' are the coordinates where the parabola intersects its axis of symmetry. In our problem, by comparing the given equation \[(x+\frac{1}{2})^2=4(y-1)\]with the standard form, we identify the vertex at \((-\frac{1}{2}, 1)\).

• The vertex can be found by inspecting the equation and comparing it with the standard form.
• The coordinates \((h, k)\) provide us with precise location details.
• Understanding this helps in easily sketching the graph as it provides the pivotal center.

Focus

The focus of a parabola is a very interesting concept. It refers to a fixed point that is used in the construction of the parabolic curve. Every point on the parabola is equidistant from the focus and the directrix, another fixed line.
The focus impacts the shape and direction of the parabola. For a parabola in the form of \((x-h)^2 = 4p(y-k)\), the focus is located at the coordinates \((h, k+p)\) since the parabola opens either upwards or downwards depending on the sign of 'p'. For our given equation \[(x+\frac{1}{2})^2=4(y-1)\],the focus is determined by moving p units along the y-axis from the vertex, due to the vertical opening.
The value of 'p' is 1, so the focus, in this case, is at \((-\frac{1}{2}, 2)\).

• The focus lies directly along the axis of symmetry of the parabola.
• It is essential for defining the precise curvature of the parabolic arc.
• The focus helps to understand how opening direction and width are defined.

Directrix

The directrix is a fundamental line associated with each parabola. It is not a point but a horizontal line (for vertical parabolas) that sits opposite the focus. It serves as a reference line for measuring distances.
The directrix is essential when determining the steepness and width of a parabolic curve. For a parabola described by \((x-h)^2 = 4p(y-k)\), the directrix is the line y = k - p.
In our scenario, with\[(x+\frac{1}{2})^2=4(y-1)\],our directrix is calculated as y = 1 - 1 = 0.

• The directrix helps set the boundary conditions for the parabola.
• It provides contrast with the focus to help define the exact shape of the parabola.
• Understanding the directrix is vital for sketching accurate parabolas.