Problem 66 - Thomas Calculus

Question

Find the center, foci, vertices, asymptotes, and radius, as appropriate, of the conic sections in Exercises \(57-68 .\) $$ x^{2}-y^{2}+4 x-6 y=6 $$

Step-by-Step Solution

Answer

Center: (-2, -3), Vertices: (-3, -3) and (-1, -3), Foci: (-2±√2, -3), Asymptotes: y = x-1 and y = -x-5.

Step 1: Rewrite in Standard Form

To begin solving the problem, we must rewrite the given equation in a standard form of a hyperbola. Our equation is: \(x^{2} - y^{2} + 4x - 6y = 6\). We need to complete the square for both \(x\) and \(y\) terms.

Step 2: Complete the Square

Rewrite the equation by rearranging and grouping: \((x^{2} + 4x) - (y^{2} + 6y) = 6\).Complete the square for \(x\):1. Take \(4/2 = 2\), square it to get \(4\).2. Add and subtract \(4\) inside the parentheses: \((x^{2} + 4x + 4 - 4)\).3. This becomes \((x + 2)^{2} - 4\).Complete the square for \(y\):1. Take \(-6/2 = -3\), square it to get \(9\).2. Add and subtract \(9\) inside the parentheses: \(-(y^{2} + 6y + 9 - 9)\).3. This becomes \(-(y + 3)^{2} + 9\).

Step 3: Simplify the Equation

Substitute the completed squares back into the equation: \((x + 2)^2 - 4 - ((y + 3)^2 - 9) = 6\).Combine constants: \((x + 2)^2 - (y + 3)^2 = 6 + 4 - 9\).Simplify to get \((x + 2)^2 - (y + 3)^2 = 1\). This is the standard form of a hyperbola centered at \((-2, -3)\).

Step 4: Identify the Center and Axes

The equation \((x + 2)^2 - (y + 3)^2 = 1\) represents a hyperbola centered at \((-2, -3)\) with its transverse axis along the \(x\)-direction.

Step 5: Find Vertices and Foci

For a hyperbola \((x - h)^2/a^2 - (y - k)^2/b^2 = 1\), the vertices are \((h \pm a, k)\).Here, \(h = -2\), \(k = -3\), \(a^2 = 1\) thus \(a = 1\).Vertices: \((-2 \pm 1, -3) = (-3, -3)\) and \((-1, -3)\).The foci are \((h \pm c, k)\) where \(c^2 = a^2 + b^2\). Since \(b = 1\), \(b^2 = 1\), \(c^2 = 2\), thus \(c = \sqrt{2}\).Foci: \((-2 \pm \sqrt{2}, -3)\).

Step 6: Determine Asymptotes

The asymptotes of the hyperbola \((x - h)^2/a^2 - (y - k)^2/b^2 = 1\) are given by the equations: \[ y - k = \pm \frac{b}{a}(x - h) \]Substituting the known values: \[ y + 3 = \pm 1 (x + 2) \]Thus, the asymptotes are \(y = x - 1\) and \(y = -x - 5\).

Key Concepts

Conic Sections

A conic section is a curve obtained by intersecting a cone with a plane. These shapes include circles, ellipses, parabolas, and hyperbolas. Each type has a unique equation and geometric properties.

Depending on the angle and position of the plane relative to the cone, the result is a different conic section.

• Circles and ellipses are formed when the plane cuts through only one nappe of the cone.
• Parabolas occur when the plane is parallel to a generating line of the cone.
• Hyperbolas are formed when the plane intersects both nappes of the cone.

Hyperbolas, like the one in this exercise, have two branches. They open either horizontally or vertically depending on their standard form equation, which is touched upon in later sections.

Completing the Square

Completing the square is a method used to convert a quadratic equation into a form that reveals key characteristics like vertex or center of a conic section.In our exercise, we had the equation: \( x^2 - y^2 + 4x - 6y = 6 \).

To find its hyperbolic nature, we need to rewrite it in standard form by completing the square for both \( x \) and \( y \). This involves:

• Reorganizing the terms: \((x^2 + 4x) - (y^2 + 6y) = 6\).
• Completing the square for \(x\): Add and subtract 4 within the equation: \((x + 2)^2 - 4\).
• Completing the square for \(y\): Add and subtract 9: \(-(y + 3)^2 + 9\).

This targeted manipulation leads us to the standard form of the hyperbola, which clearly highlights the conic's center and helps to identify the axes.

Standard Form of a Hyperbola

The standard form of a hyperbola is crucial for understanding its features, such as its orientation, center, and asymptotes. For a hyperbola centered at \((h, k)\) oriented horizontally, the equation is:\[(x - h)^2/a^2 - (y - k)^2/b^2 = 1\]If oriented vertically, the equation is:\[(y - k)^2/a^2 - (x - h)^2/b^2 = 1\]In our exercise, after completing the square, the hyperbola's equation is:\[(x + 2)^2 - (y + 3)^2 = 1\]This indicates a horizontal orientation centered at \((-2, -3)\).

The values of \(a^2\) and \(b^2\) give insights into the hyperbola's shape. Here, both \(a^2\) and \(b^2\) are 1, indicating a symmetrically shaped hyperbola around its center.

Vertices of a Hyperbola

Vertices are specific points on the hyperbola that lie on the axes and represent the 'widest' part of the curve. For a hyperbola in standard form, if it is horizontally aligned like ours, the vertices can be found at:

• \((h \pm a, k)\)

For our hyperbola, centered at \((-2, -3)\) with \(a = 1\), the vertices are located at:

• \((-3, -3)\)
• \((-1, -3)\)

These vertices mark the closest approach of the hyperbola branches along the transverse axis.

Knowing the vertices allows us to visualize the shape and orientation of the hyperbola more effectively.

Asymptotes of a Hyperbola

Asymptotes are lines that a hyperbola approaches but never intersects. For hyperbolas, they aid in understanding the overall shape and direction of the branches.The equation for the asymptotes of a hyperbola in its standard form is:\[y - k = \pm \frac{b}{a}(x - h)\]For our equation \((x + 2)^2 - (y + 3)^2 = 1\), the asymptotes can be derived as follows:

• Substituting the values: \(y + 3 = \pm 1(x + 2)\)
• This simplifies to: \(y = x - 1\) and \(y = -x - 5\)

The asymptotes provide a visual framework for sketching the hyperbola, allowing us to approximate its curvature and overall direction. They are key to understanding the infinite extent of the hyperbola's branches.