Problem 65 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The hyperbola has center (1,2), vertices at (4,2) and (-2,2), foci at \((1 ± 3\sqrt{2}, 2)\), and asymptotes \(y = x+1\) and \(y = -x+3\).

Step 1: Rewrite in Standard Form

We start by rearranging and completing the square to rewrite the given equation in standard form. The given equation is \(x^2 - y^2 - 2x + 4y = 4\). First, complete the square for both \(x\) and \(y\):\[(x^2 - 2x) - (y^2 - 4y) = 4\]For \(x\): Complete the square:\[(x^2 - 2x) = (x-1)^2 - 1\]For \(y\): Complete the square:\[(y^2 - 4y) = (y-2)^2 - 4\]Substitute these back into the equation:\[(x-1)^2 - 1 - ((y-2)^2 - 4) = 4\]Simplify:\[(x-1)^2 - (y-2)^2 = 9\]Now the equation is in standard form: \((x-1)^2 - (y-2)^2 = 3^2\).

Step 2: Identify the Type of Conic Section

The equation \((x-1)^2 - (y-2)^2 = 3^2\) represents a hyperbola because it is in the form of \((x-h)^2/a^2 - (y-k)^2/b^2 = 1\), where there is a subtraction between the \(x\) and \(y\) terms.

Step 3: Find the Center

The standard form \((x-h)^2/a^2 - (y-k)^2/b^2 = 1\) shows that the center of the hyperbola is \((h,k)\). Therefore, the center is \((1,2)\).

Step 4: Determine the Vertices

The vertices of the hyperbola are located \(a\) units from the center along the direction of the \(x\)-axis. Here, \(a = 3\). Thus, the vertices are:\[(1 + 3, 2) = (4, 2)\]\[(1 - 3, 2) = (-2, 2)\]

Step 5: Calculate the Foci

For a hyperbola, the relationship is \(c^2 = a^2 + b^2\) where \(c\) is the distance from the center to the foci. Here, \(a = 3\) and \(b = 3\) (since both are implied and equal for this hyperbola), hence:\[c^2 = 3^2 + 3^2 = 18\]\(c = \sqrt{18} = 3\sqrt{2}\)The foci are located \(c\) units from the center:\[(1 + 3\sqrt{2}, 2)\] and \[(1 - 3\sqrt{2}, 2)\]

Step 6: Determine the Asymptotes

For a hyperbola in the standard form \((x-h)^2/a^2 - (y-k)^2/b^2 = 1\), the equations of the asymptotes are:\[y - k = \pm \frac{b}{a}(x - h)\]With \(a = 3\) and \(b = 3\), and \((h,k) = (1,2)\), the asymptotes are:\[y - 2 = \pm 1(x - 1)\]This simplifies to:\[y = x + 1\] and \[y = -x + 3\]

Key Concepts

Standard Form

A hyperbola's standard form helps us identify its important features, such as its center, vertices, and axes. Converting the general quadratic equation into standard form is crucial for understanding its structure. For a hyperbola, the standard form is often expressed as:\[ (x-h)^2/a^2 - (y-k)^2/b^2 = 1 \]where \(h\) and \(k\) represent the coordinates of the hyperbola's center, and \(a\) and \(b\) are the distances from the center to the vertices and co-vertices respectively. Achieving this form requires careful manipulation, often through completing the square.

Completing the Square

Completing the square is a technique used to make a quadratic expression easier to work with, typically transforming it into a perfect square trinomial. It is especially handy when rewriting conic sections into a standard form. Here's how it's done:

• Group the \(x\) and \(y\) terms.
• Focus on each set of terms: add and subtract the square of half the coefficient of the linear term within the quadratic expression.
• This process forms a perfect square trinomial, which can then easily be expressed in squared binomial form.

For example, converting \(x^2 - 2x\) becomes \((x-1)^2 - 1\) when completing the square.

Conic Sections

Conic sections are the curves obtained by intersecting a plane with a double-napped cone. These sections include ellipses, parabolas, circles, and hyperbolas. Each has a unique equation form associated with it based on its geometric properties. Hyperbolas are distinct because their equation includes a subtraction sign between their squared terms. This distinction helps easily identify hyperbolas when given their equations. In the form \((x-h)^2/a^2 - (y-k)^2/b^2 = 1\), the hyperbola is recognized by the separating minus sign and its graph typically consists of two separate curves.

Vertices of Hyperbola

The vertices of a hyperbola are key features that define its shape. They are the closest points on each branch of the hyperbola to the center. To find the vertices of a hyperbola given by \((x-h)^2/a^2 - (y-k)^2/b^2 = 1\), we use the value of \(a\):

• One vertex will be at \((h+a, k)\).
• The other vertex will be located at \((h-a, k)\).

In our example involving \(a = 3\), the vertices calculated are at \((4, 2)\) and \((-2, 2)\). The distance from the center along the x-axis helps sketch the approximate shape of the hyperbola.

Foci of Hyperbola

The foci of a hyperbola are significant points located along the axis of symmetry. They lie outside the branches and help explain the geometry of the hyperbola, primarily concerning the way distances add up. We find the foci using the formula:\[ c^2 = a^2 + b^2 \]where \(c\) is the distance from the center to each focus. For our example:

• Calculate \(c = \sqrt{18} = 3\sqrt{2}\).
• The foci are then found at \((1 + 3\sqrt{2}, 2)\) and \((1 - 3\sqrt{2}, 2)\).

The foci's positioning affects the curvature and direction of the branches.

Asymptotes of Hyperbola

Asymptotes of a hyperbola are straight lines that the curve approaches but never touches. They provide guidance for sketching hyperbolas elegantly, helping to understand their breadth and direction. For hyperbolas, the asymptotes are determined by:\[ y - k = \pm \frac{b}{a}(x - h) \]In this example, with \(a = b = 3\) and center \((1, 2)\), the asymptotes become:

• \[ y = x + 1 \]
• \[ y = -x + 3 \]

These equations indicate the slope and intercepts of the lines that guide the hyperbola's open arms.