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

Step 1: Rearrange the Equation

Start by rearranging the given equation to separate the terms involving the same variable: \[x^2 + 4x - y^2 - 6y = 6\]

Step 2: Complete the Square for x

To complete the square for terms involving \(x\):- Consider \(x^2 + 4x\).- Take half of the coefficient of \(x\), which is \(4/2 = 2\), square it, and add and subtract it: \[(x^2 + 4x + 4) - 4\]- This becomes: \[(x+2)^2 - 4\]

Step 3: Complete the Square for y

To complete the square for terms involving \(y\):- Consider \(-y^2 - 6y\).- Take half of the coefficient of \(y\), which is \(-6/2 = -3\), square it, and add and subtract it: \[(-(y^2 + 6y + 9) + 9)\]- This becomes:\[-(y+3)^2 + 9\]

Step 4: Rewrite the Equation

Substitute the completed squares back into the equation:\[(x+2)^2 - 4 - (y+3)^2 + 9 = 6\]Combine the constants:\[(x+2)^2 - (y+3)^2 = 1\]

Step 5: Identify the Conic Section

The equation \((x+2)^2 - (y+3)^2 = 1\) is the standard form of a hyperbola centered at \((-2, -3)\) with the transverse axis parallel to the x-axis.

Step 6: Calculate the Vertices

For a hyperbola in the form \((x-h)^2 - (y-k)^2 = 1\),- The vertices are located at \((h+a, k)\) and \((h-a, k)\).- Here, \(a = 1\), so the vertices are at \((-2+1, -3) = (-1, -3)\) and \((-2-1, -3) = (-3, -3)\).

Step 7: Find the Foci

For a hyperbola in this form, the foci are located \(c\) units away from the center along the transverse axis, where \(c^2 = a^2 + b^2\) and \(b^2 = 1\).- Here, \(a = 1\), \(b = 1\), thus \(c^2 = 2\) and \(c = \sqrt{2}\).- The foci are \((-2 + \sqrt{2}, -3)\) and \((-2 - \sqrt{2}, -3)\).

Step 8: Determine the Asymptotes

The asymptotes of a hyperbola \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\) are given by the equations:\[y-k = \pm \frac{b}{a}(x-h)\]For this hyperbola:- \(b = 1\), \(a = 1\), hence:- The equations are:\[y + 3 = \pm (x + 2)\]Or\[y = x - 1\] and \[y = -x - 5\]

Key Concepts

Conic Sections

Conic sections are shapes created by the intersection of a cone with a plane. These intersections can result in four different shapes: circles, ellipses, parabolas, and hyperbolas. Each conic section has unique properties and equations that describe its geometry and characteristics.
For the problem we are discussing, we are looking at a hyperbola, which is one type of conic section. Hyperbolas have two separate curves called branches, and they differ from ellipses and parabolas because they open outwards, rather than closing in on themselves.

• A hyperbola has two axes: the transverse axis, which lies along the direction the branches open, and the conjugate axis, which is perpendicular to it.
• The center of a hyperbola is the midpoint between its vertices and the foci are points located inside each branch along the transverse axis.
• Understanding the layout of a hyperbola is crucial for finding elements such as vertices, foci, and asymptotes from its equation, as shown in the exercise example.

Completing the Square

Completing the square is a mathematical technique used to rewrite quadratic equations in a form that makes them easier to analyze or solve. It is especially useful for identifying conic sections from equations that are not immediately recognizable in standard form.
To complete the square, we manipulate the terms in a quadratic expression, adding and subtracting specific numbers. This method "completes" a perfect square trinomial:

• For example, in the equation given in the exercise, completing the square involved reorganizing the terms for both variables \(x\) and \(y\). This was done by finding the value needed to create a square, adding and subtracting this value within the equation.
• For the \(x\) terms \(x^2 + 4x\), adding and subtracting 4 results in \((x+2)^2 - 4\).
• For the \(y\) terms \(-y^2 - 6y\), adding and subtracting 9 results in \(-(y+3)^2 + 9\).

This technique ultimately allows us to transform the original equation into a standard form for a hyperbola, enabling further analysis.

Asymptotes

Asymptotes are straight lines that a curve approaches but never actually reaches. In the context of hyperbolas, asymptotes play a critical role in outlining the direction and steepness of the branches of the hyperbola.
In our example, after transforming the equation into standard form, the asymptotes help define the shape of the hyperbola. Hyperbolas always have two asymptotes, which intersect at the center of the hyperbola and extend outward.

• For a hyperbola expressed as \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), the equations for the asymptotes are \(y-k = \pm \frac{b}{a}(x-h)\).
• For the specific case in the exercise, with \(a = 1\) and \(b = 1\), the asymptotes are described by the lines \(y = x - 1\) and \(y = -x - 5\).

These asymptotes guide the general angle and arrangement of the hyperbola's branches.