Problem 78 - A Graphical Approach to College Algebra

Question

Write the equation in standard form for a hyperbola centered at ( \(h, k\) ). Identify the center and vertices. \(y^{2}+4 y-x^{2}+2 x=6\)

Step-by-Step Solution

Answer

The standard form is \( \frac{(y+2)^2}{11} - \frac{(x-1)^2}{11} = 1 \) with center at (1, -2) and vertices at (1, -2±√11).

Step 1: Rearrange the Equation

Start by rearranging the given equation to align the terms involving the same variable:\[ y^2 + 4y - x^2 + 2x = 6 \]Group the terms with respect to the same variables (i.e., group \(y\) terms and \(x\) terms separately):\[ (y^2 + 4y) - (x^2 - 2x) = 6 \]

Step 2: Complete the Square

Complete the square for each group of terms. For the \(y\) terms:Take the coefficient of \(y\), divide by 2, and square it:\[ \left(\frac{4}{2}\right)^2 = 4 \]Add and subtract 4 within the \(y\) terms:\[ (y^2 + 4y + 4 - 4) - (x^2 - 2x) = 6 \]This simplifies to:\[ ((y + 2)^2 - 4) - (x^2 - 2x) = 6 \]Now, for the \(x\) terms:Take the coefficient of \(x\), divide by 2, and square it:\[ \left(\frac{-2}{2}\right)^2 = 1 \]Add and subtract 1 within the \(x\) terms:\[ ((y + 2)^2 - 4) - (x^2 - 2x + 1 - 1) = 6 \]This simplifies to:\[ (y + 2)^2 - (x - 1)^2 - 4 - 1 = 6 \]

Step 3: Simplify the Equation

Simplify the equation by combining constants:\[ (y + 2)^2 - (x - 1)^2 - 5 = 6 \]Add 5 to both sides:\[ (y + 2)^2 - (x - 1)^2 = 11 \]

Step 4: Write in Standard Form

The standard form of a hyperbola with center \((h, k)\) is: \[ \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \]Comparing with the equation:\[ (y + 2)^2 - (x - 1)^2 = 11 \]We get:\[ \frac{(y+2)^2}{11} - \frac{(x-1)^2}{11} = 1 \]

Step 5: Identify the Center and Vertices

The center of this hyperbola is at \((h, k) = (1, -2)\) because it matches the transformed standard form.The vertices of the hyperbola are determined by the distance \(a\) along the \(y\)-axis for this vertical hyperbola. From the equation, \(a^2 = 11\), so \(a = \sqrt{11}\). Thus, the vertices are at \( (1, -2 \pm \sqrt{11}) \).

Key Concepts

Completing the Square

Completing the square is an essential technique in algebra used to rewrite a quadratic equation into a simpler form. It's particularly useful for transforming equations of conics, like hyperbolas, into their standard forms. By completing the square, we can reveal the true shape of the curve and make it easier to analyze.

Let's consider a quadratic term, such as the given \[(y^2 + 4y)\]. To complete the square:

• Take the coefficient of the linear term (i.e., the term with the variable to the first power), which in this case is 4 from \(y\), and divide by 2. This gives us 2.
• Then, square the result from above, which is \(2^2 = 4\).
• Add and subtract this squared term within the parenthesis to complete the square and keep the equation balanced.

This manipulates our quadratic expression into \((y+2)^2 - 4\), an easily recognizable perfect square trinomial. A similar process is applied to the \(x\) terms to rewrite \((x^2 - 2x)\) as \((x-1)^2 -1\).By practicing completing the square, it allows the transformation of quadratic expressions into easily interpretable formats.

Standard Form

The standard form of a hyperbola simplifies understanding its properties and quickly identifies its center and alignment. The standard equation of a hyperbola is \[\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\] for a vertical hyperbola.

Based on the original exercise, we've managed to manipulate and transform \[(y+2)^2 - (x-1)^2 = 11\] into \[\frac{(y+2)^2}{11} - \frac{(x-1)^2}{11} = 1\]. This format offers clean insight into key geometric characteristics:

• The terms \( (y-k) \) and \( (x-h) \) confirm the position of the center \((h, k)\).
• The values of \(a^2\) and \(b^2\) reflect the dimensions of the box the hyperbola forms along its axes.

Standard form streamlines processes when analyzing conics, as values for center and talents, like intercepts and asymptotes, can be identified swiftly.

Center and Vertices

The center is the reference point from which the symmetry of the hyperbola originates, while the vertices reveal points of greatest or least distance along the transverse axis.

For the hyperbola equation in standard form \[\frac{(y+2)^2}{11} - \frac{(x-1)^2}{11} = 1\], the values \(h = 1\) and \(k = -2\) designate the center of the hyperbola. Thus, the center point is \((1, -2)\).

Vertices are determined along the axis where the first term variable resides. Here, since \(y\) is the first variable, it indicates a vertical hyperbola. The distance \(a = \sqrt{11}\) quantifies how far vertices are from the hyperbola's center along the vertical axis. Therefore, the vertices occur at positions \((1, -2+\sqrt{11})\) and \((1, -2-\sqrt{11})\).

Understanding the center and vertices is critical as it outlines the primary shape and orientation of a hyperbola within the coordinate plane.