Problem 31 - 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. $$x^{2}+16=4\left(y^{2}+2 x\right)$$

Step-by-Step Solution

Answer

The equation represents a parabola with vertex (4, 0), focus (4, 1), and directrix y = -1.

Step 1: Expand and Rearrange the Equation

Start with the given equation: \(x^2 + 16 = 4(y^2 + 2x)\). Distribute the 4 on the right side to expand it: \(x^2 + 16 = 4y^2 + 8x\). Rearrange this to group like terms on one side: \(x^2 - 8x + 16 = 4y^2\).

Step 2: Complete the Square for the x-Terms

To complete the square for \(x^2 - 8x\), take half of the coefficient of \(x\) which is -8 to get -4, then square it to get 16. Thus, the equation becomes \((x-4)^2 = 4y^2\).

Step 3: Compare with Standard Forms of Conics

The equation \((x-4)^2 = 4y^2\) suggests a comparison with the standard form of a parabola, which is typically \((x-h)^2 = 4p(y-k)\). In this equation, \(h = 4\) and \(k = 0\).

Step 4: Identify the Vertex, Focus, and Directrix of the Parabola

The vertex of the parabola is located at \((h, k) = (4, 0)\). From the form \(4p = 4\), we find that \(p = 1\). This implies that the focus, which is \((h, k+p) = (4, 1)\), is found one unit along the axis of symmetry of the parabola. The directrix, being \(y = k - p = -1\), is one unit below the vertex.

Step 5: Final Check and Sketch

Double-check the calculations and interpretations. The equation represents a parabola, verifying that it opens up based on \((x-h)^2 = 4p(y-k)\) being positive. Sketch the parabola on a coordinate plane with vertex \((4, 0)\), focus at \((4, 1)\), and directrix at \(y = -1\).

Key Concepts

Completing the Square

Completing the square is a valuable technique for transforming quadratic equations into a form that reveals key features of conic sections, such as parabolas. In this process, we focus on terms containing a specific variable, typically quadratics like \(x^2\). Say we have an expression: \(x^2 + bx\). The goal is to turn it into \((x - d)^2\). Here's how it works:

• Identify the coefficient of \(x\), which is \(b\) in \(x^2 + bx\).
• Take half of \(b\), that is \(\frac{b}{2}\).
• Square \(\frac{b}{2}\) to get \(\left(\frac{b}{2}\right)^2\).
• Add and subtract \(\left(\frac{b}{2}\right)^2\) within the expression to maintain balance.

For example, in the problem: \(x^2 - 8x\), half of -8 is -4. Squaring -4 gives 16, leading us to the expression \((x - 4)^2\). Completing the square simplifies identifying certain properties of the equation, such as whether it forms a parabola.

Parabola

Parabolas are unique shapes found in algebra that emerge from certain quadratic equations. A parabola's equation can often be expressed in vertex form: \((x-h)^2 = 4p(y-k)\). Here's what this means:

• The vertex of the parabola is the point \((h, k)\), which serves as the peak or trough of the parabola.
• The variable \(p\) determines how wide or narrow the parabola is, as well as its orientation (upwards or downwards).
• The symmetry of the parabola is around the line \(x = h\), meaning the shape is mirrored perfectly on either side of \(h\).

In the completed square equation \((x-4)^2 = 4(y^2)\), it directly correlates to a parabola that opens upwards. The transformation of the original equation to this standard form helps us see its orientation and symmetry quickly.

Vertex and Focus

The vertex and focus are critical features that precisely describe the parabola's shape and position. Here's how they relate:

• The vertex, \((h, k)\), is a pivotal point. For the given parabola, it is at \((4, 0)\).
• From the expression \(4p = 4\), solve for \(p = 1\), giving the distance from the vertex to the focus and the directrix.
• The focus of the parabola \((h, k + p)\) is at \((4, 1)\), indicating the point toward which the parabola tucks in.
• The directrix is a horizontal line opposite to the focus at \(y = k - p = -1\). Points on the parabola are equidistant from this line and the focus.

Understanding these relationships forms a comprehensive picture of how parabolas behave, helping in sketching and analyzing them.