Problem 33 - Precalculus with Limits

Question

In Exercises 33-46, find the vertex, focus, and directrix of the parabola, and sketch its graph. \(y = \frac{1}{2}x^2\)

Step-by-Step Solution

Answer

The vertex of the parabola is at (0, 0). The focus is at the point (0, 2) and the equation of the directrix is \(y = -2\). The sketch of the graph depicts a parabola that opens upwards with the vertex at the origin, the focus above the x-axis, and the directrix below the x-axis.

Step 1: Identify the vertex

Since the equation \(y = \frac{1}{2}x^2\) is in the form of standard parabolic equation \(y = ax^2\), the vertex is at the origin (0,0). Therefore, the vertex of the parabola is (0, 0).

Step 2: Find the Focus

The formula to find the focus of a parabola given in the standard form is (h,k+1/4a) where h and k are the vertex and a is the coefficient of \(x^2\). Here, the coefficient a is equal to 1/2. Therefore, the focus will be at (0,2).

Step 3: Identify the Directrix

The equation for the directrix of a parabola in the standard form is \(y = -\frac{1}{4a}\). Substituting the given value of a (1/2), the equation of the directrix becomes \(y = -2\). Therefore, the directrix of the parabola is \(y = -2\).

Step 4: Sketch the graph

Start by identifying the vertex at (0,0) on the graph, then plot the focus at the point (0,2). Draw the directrix line at \(y = -2\). Finally, sketch the parabola using these points and line. The parabola will open upwards, as a is positive.

Key Concepts

Vertex of a Parabola

Understanding the vertex of a parabola is crucial in graphing and analyzing the parabolic shape. The vertex is the highest or lowest point on the parabola, depending on whether it opens upwards or downwards. In our exercise, the parabola given by the equation \(y = \frac{1}{2}x^2\) has a vertex at the origin (0,0).

This result can be determined by recognizing that the standard form of a parabola \(y = ax^2 + bx + c\) has a vertex at \((h, k)\), where \(h = -\frac{b}{2a}\) and \(k\) is the value of the equation for \(x = h\). Here, since \(b\) and \(c\) are both 0, the vertex defaults to (0,0), which is also confirmed through completing the square method, if necessary. Knowing the vertex helps us graph the parabola accurately and understand its properties better.

Focus of a Parabola

The focus of a parabola is a fixed point inside the curve from which each point on the parabola is equidistant to a corresponding point on the directrix. For the parabola \(y = \frac{1}{2}x^2\), the focus lies at a specific point that we can determine using the formula \((h, k + \frac{1}{4a})\), where \(a\) is the coefficient of \(x^2\) in the parabolic equation.

In the given parabola, \(a\) is \(\frac{1}{2}\), hence the focus is at \((0, 0 + \frac{1}{4 \cdot \frac{1}{2}}) = (0, 2)\). It is located along the axis of symmetry, which in this case is the y-axis, and exactly 2 units above the vertex. The focus not just helps us in the plotting but also in understanding the reflective properties of parabolas, especially in the context of optics and physics.

Directrix of a Parabola

for the shape of the parabola and a reference for the locus definition of a parabola. Whether you're into geometry or you're designing a satellite dish, understanding the directrix is key.

Graphing Parabolas

Graphing a parabola involves plotting the vertex, focus, and directrix and then sketching the curve that fits these elements. With the parabola \(y=\frac{1}{2}x^2\), we start with the vertex at (0,0). Next, we plot the focus at (0,2) and draw the horizontal line for the directrix at \(y=-2\).

Armed with these points, we proceed to graph the parabola. This particular parabola opens upwards, as indicated by the positive coefficient of \(x^2\). The curve is symmetric with respect to the line x = 0, which means if we plot a point with coordinates (x, y), we should also plot (-x, y) to reflect its symmetry. The distance between any point on the parabola and the focus is the same as from that point to the directrix, this principle guides the curvature. Using these relationships ensures an accurate representation of the parabola.