Problem 30 - Precalculus Mathematics for Calculus

Question

Complete the square to determine whether the equation represents an ellipse, a parabola, a hyperbola, or a degenerate conic. If the graph is an ellipse, find the center, foci, vertices, and lengths of the major and minor axes. If it is a parabola, find the vertex, focus, and directrix. If it is a hyperbola, find the center, foci, vertices, and asymptotes. Then sketch the graph of the equation. If the equation has no graph, explain why. $$4 x^{2}-4 x-8 y+9=0$$

Step-by-Step Solution

Answer

The equation represents an upward-opening parabola with vertex \((\frac{1}{2}, 1)\), focus \((\frac{1}{2}, \frac{3}{2})\), and directrix \(y = \frac{1}{2}\).

Step 1: Rearrange the terms

Start by rearranging the given equation to group the x-terms together and isolate the constant on one side. The given equation is \(4x^2 - 4x - 8y + 9 = 0\). Rearrange as \(4x^2 - 4x = 8y - 9\).

Step 2: Factor out a leading coefficient

In the x-term \(4x^2 - 4x\), factor out the 4 to simplify the terms for completing the square. This gives us \(4(x^2 - x) = 8y - 9\).

Step 3: Complete the square

Take the expression inside the parenthesis \(x^2 - x\) and complete the square. To do this, add and subtract \(\left(\frac{1}{2}\right)^2\) (which is \(\frac{1}{4}\)) inside the parenthesis:\[x^2 - x + \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2\]This becomes:\[x^2 - x + \frac{1}{4} - \frac{1}{4} = \left(x - \frac{1}{2}\right)^2 - \frac{1}{4}\].

Step 4: Insert the completed square into the equation

Replace \(x^2 - x\) in the equation with \(\left(x - \frac{1}{2}\right)^2 - \frac{1}{4}\):\[4\left(\left(x - \frac{1}{2}\right)^2 - \frac{1}{4}\right) = 8y - 9\]Simplify to:\[4\left(x - \frac{1}{2}\right)^2 - 1 = 8y - 9\].

Step 5: Solve for y

Rearrange the equation to solve for y:\[8y = 4\left(x - \frac{1}{2}\right)^2 + 8\]Divide all terms by 8 to isolate y:\[y = \frac{1}{2}(x - \frac{1}{2})^2 + 1\].

Step 6: Identify the conic type and components

The equation \(y = \frac{1}{2}(x - \frac{1}{2})^2 + 1\) matches the form of a parabola \(y = a(x-h)^2 + k\), indicating that this is a parabola. It opens upwards with vertex at \((\frac{1}{2}, 1)\), and parameter \(\frac{1}{4a} = \frac{1}{4 \times \frac{1}{2}} = \frac{1}{2}\). The focus is at \((\frac{1}{2}, \frac{3}{2})\), and the directrix is the line \(y = \frac{1}{2}\).

Step 7: Sketch the graph

The parabola opens upwards from its vertex at \((\frac{1}{2}, 1)\). The focus at \((\frac{1}{2}, \frac{3}{2})\) is inside the parabola, and the directrix, \(y = \frac{1}{2}\), is a horizontal line below the vertex. Draw the parabola to illustrate these points.

Key Concepts

Conic Sections

Conic sections are curves obtained by slicing a cone with a plane. Depending on the angle and position of the plane, the intersections produce different shapes. These shapes include:

• Circle - A special case of an ellipse, appearing when the cutting plane is perpendicular to the cone's axis.
• Ellipse - Formed when the plane cuts the cone at an angle, but doesn't intersect the base.
• Parabola - Created when the plane is parallel to a generatrix (slant height) of the cone.
• Hyperbola - Occurs when the plane cuts through both nappes (the top and bottom) of the cone.

These conic sections are not just mathematical abstractions; they appear naturally in various scientific phenomena. For instance, planets trace elliptical orbits around the sun, and parabolas describe projectile paths. Understanding these curves provides insights into a galaxy of applications ranging from astronomy to engineering.

In this exercise, the task was to identify which type of conic the given equation represents and to identify its characteristics.

Parabola

A parabola is a curve that mirrors itself and is shaped like an open bowl. Every parabola has a unique property—it remains equidistant from a point called the "focus" and a line termed the "directrix." This intriguing relationship produces the characteristic U-shaped form.

Mathematically, a parabola can be expressed in the standard form if it opens upwards or downwards: \[ y = a(x-h)^2 + k \] where:

• (h, k) is the vertex of the parabola.
• a affects the width and the direction of the opening (upwards if a is positive, downwards if negative).

The equation helps visualize how the parabola behaves on the coordinate plane. This exercise demonstrated how completing the square transforms the equation into this form, revealing its parabolic nature.

Vertex and Focus of a Parabola

Identifying the vertex and focus is crucial to sketching a parabola and understanding its orientation. The vertex is the turning point of the parabola, where it changes direction. For the equation \[ y = \frac{1}{2}(x-\frac{1}{2})^2 + 1 \] the vertex is \[ (\frac{1}{2}, 1) \].

Meanwhile, the focus is a special point inside the parabola that, together with the directrix, defines the parabola's shape and position. For this parabola, the focus is at \[ (\frac{1}{2}, \frac{3}{2}) \]. The distance from the vertex to the focus is the same as from the vertex to the directrix, and it's calculated using the formula \[ \left(\frac{1}{4a}\right) \].

With the vertex and focus identified, along with the directrix, which is the line \[ y = \frac{1}{2} \], the entire structure of the parabola is easily constructed, allowing for a clear sketch and understanding of its properties.