Problem 60 - Thomas Calculus

Question

Find the center, foci, vertices, asymptotes, and radius, as appropriate, of the conic sections in Exercises \(57-68 .\) $$ y^{2}-4 y-8 x-12=0 $$

Step-by-Step Solution

Answer

Vertex: (-2, 2); Focus: (0, 2); Directrix: x = -4; Parabola.

Step 1: Identify the type of conic section

The given equation is quadratic in nature. Since it contains both squared and linear terms in one variable and only a linear term in the other variable, it is a rotated conic section. Our first task is to identify whether it's a parabola, hyperbola, ellipse, or circle based on its form. In this case, it resembles a parabola, as only one variable is squared.

Step 2: Complete the square

Rewrite the given equation to make the squared term complete. Rearrange it to focus on the quadratic part: \[ (y^2 - 4y) = 8x + 12 \] Now, complete the square for \( y^2 - 4y \):\[ y^2 - 4y = (y-2)^2 - 4 \] Therefore, the equation becomes: \[ (y-2)^2 - 4 = 8x + 12 \]

Step 3: Simplify the equation

Move the \(-4\) from the left side to the right to simplify:\[ (y-2)^2 = 8x + 16 \] Simplify further to get: \[ (y-2)^2 = 8(x + 2) \] This is the standard form of a parabola with vertex \( (h, k) \).

Step 4: Identify the vertex

From the standard parabola form \((y-k)^2 = 4p(x-h)\), we have \((y-2)^2 = 8(x + 2)\). This shows the vertex is at \( (-2, 2) \).

Step 5: Determine the focus

In the standard parabola form \((y-k)^2 = 4p(x-h)\), we have that \(4p = 8\), thus \(p = 2\).The parabola opens to the right, so the focus is \(p\) units from the vertex along the x-axis at \((-2 + 2, 2) = (0, 2)\).

Step 6: Identify the directrix

The directrix is \(p\) units to the left of the vertex. Since \(p = 2\) and the vertex is at \(-2, 2\), the directrix is the line \(x = -2 - 2 = -4\).

Key Concepts

Parabola

A parabola is one of the main types of conic sections that you will encounter in mathematics. It's defined as the set of all points that are equidistant from a fixed point called the focus and a line called the directrix. Parabolas have a distinct U-shape or, depending on the orientation, could open sideways as well. They are characterized by having exactly one squared term in their standard form equation, such as

• (y-2)^2 = 8(x + 2)

This equation indicates a parabola that opens sideways (to the right in this case). Unlike circles or ellipses, parabolas do not have a center or closed form; instead, their shape means they continue infinitely in the direction they open. Another key characteristic of parabolas is that they possess an axis of symmetry, which is a line that cuts the parabola into two mirror-image halves.

Completing the Square

Completing the square is a crucial technique in algebra that is frequently used to transform a quadratic equation into a form that is easier to work with. This form reveals important properties about a parabola, such as its vertex. The process involves the following steps:

• Start with a quadratic expression, such as \(y^2 - 4y\).
• Rearrange and rewrite it to create a perfect square trinomial. For example, in \(y^2 - 4y\), you would add and subtract a number that allows you to write it as a perfect square: \((y-2)^2\), and thus complete the square.
• This allows the equation to be expressed as \((y-2)^2 - 4\), which reveals the vertex when rewritten in standard parabola form.

By converting the quadratic to a perfect square, you're able to clearly identify the critical elements of the parabola. This includes the vertex, which is essential for graphing and understanding the parabola's orientation and position.

Vertex Form

The vertex form of a parabola is an essential tool in understanding the properties of a conic section. When a quadratic equation is written in vertex form, it appears as:

• (y-k)^2 = 4p(x-h)

In this form,

• \((h, k)\) is the vertex of the parabola.
• The term \(p\) represents the distance from the vertex to the focus and is also used to determine the width and direction of the parabola.
• For example, in (y-2)^2 = 8(x + 2), the vertex is at \(-2, 2\) and \(p = 2\), indicating a directrix and focus placement.

Understanding the vertex form provides immediate insight into the key features of a parabola, ensuring you can easily graph and interpret its orientation. Having the equation in vertex form simplifies the process of identifying transformations such as translations or reflections that may occur as compared to the general form.