Problem 81 - Intermediate Algebra

Question

Write the equation \(36 x^{2}-25 y^{2}-72 x-100 y=964\) in standard form to show that it describes a hyperbola.

Step-by-Step Solution

Answer

The equation describes a hyperbola: \(\frac{(x-1)^2}{25} - \frac{(y+2)^2}{36} = 1\).

Step 1: Group and rearrange terms

Start by rearranging and simplifying the given equation. The equation given is \(36x^2 - 25y^2 - 72x - 100y = 964.\) Group the \(x\) terms together and the \(y\) terms together. The equation becomes: \(36x^2 - 72x - 25y^2 - 100y = 964\).

Step 2: Complete the square for x-terms

To complete the square for the \(x\)-terms, factor out the coefficient of \(x^2\) from \(36x^2 - 72x\): \(36(x^2 - 2x)\). Now, complete the square inside the parenthesis: add and subtract \(1^2\), which is \((x-1)^2 - 1\). Thus, \(36(x^2 - 2x)\) becomes \(36((x-1)^2 - 1) = 36(x-1)^2 - 36\).

Step 3: Complete the square for y-terms

Similarly, for the \(y\)-terms \(-25y^2 - 100y\), factor out \(-25\): \(-25(y^2 + 4y)\). Complete the square inside the parenthesis: add and subtract \(2^2\), which is \((y+2)^2 - 4\). Thus, \(-25(y^2 + 4y)\) becomes \(-25((y+2)^2 - 4) = -25(y+2)^2 + 100\).

Step 4: Substitute back into the equation and simplify

Now substitute the completed square forms back into the equation: \[ 36(x-1)^2 - 36 - 25(y+2)^2 + 100 = 964 \]. Combine like terms to simplify: \[ 36(x-1)^2 - 25(y+2)^2 = 964 + 36 - 100 \]. Further simplifying gives us: \[ 36(x-1)^2 - 25(y+2)^2 = 900 \].

Step 5: Divide to get the standard form of a hyperbola

Divide every term in the equation \(36(x-1)^2 - 25(y+2)^2 = 900\) by 900 to get it in standard form: \[ \frac{36(x-1)^2}{900} - \frac{25(y+2)^2}{900} = \frac{900}{900} \]. This simplifies to: \[ \frac{(x-1)^2}{25} - \frac{(y+2)^2}{36} = 1 \]. This is the standard form of a hyperbola with transverse axis along the x-axis.

Key Concepts

Completing the Square

Completing the square is a technique used to transform a quadratic expression into a perfect square trinomial. This is particularly useful to convert an equation into its standard form, especially when dealing with conic sections like parabolas, ellipses, and hyperbolas. In our hyperbola equation, we apply completing the square separately for the x and y terms.
For the x-terms, start by grouping and factoring out the coefficient of the quadratic term. Here, the x-terms are combined as: \[ 36x^2 - 72x \] Factor out 36: \[ 36(x^2 - 2x) \]Next, to complete the square, we need to transform the expression inside the parenthesis into a perfect square. Take half of the linear term coefficient (-2), which is -1, square it, resulting in 1. Add and subtract this value inside the bracket:\[ 36((x-1)^2 - 1) = 36(x-1)^2 - 36 \] Similarly, for the y-terms (after rearrangement): \[-25(y^2 + 4y) \] Factor out -25:\[-25(y^2 + 4y) \]Complete the square by taking half of 4 (giving 2), squaring it to get 4, and modifying the expression:\[-25((y+2)^2 - 4) = -25(y+2)^2 + 100 \]
These steps transform the quadratic portions of the equation, making it easier to manipulate into standard form.

Standard Form

The standard form of a conic section provides a clear and succinct way to identify and analyze its properties. For hyperbolas, the standard form of the equation aligns the hyperbola with one of the coordinate axes and identifies its center, vertices, and axes.
For hyperbolas oriented along the x-axis, the standard form looks like this:\[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \]Where **h** and **k** represent the coordinates of the center, **a** is the distance from the center to each vertex along the transverse axis, and **b** is the distance along the conjugate axis. In our example, after transforming and simplifying, we find the standard form as:\[ \frac{(x-1)^2}{25} - \frac{(y+2)^2}{36} = 1 \]This indicates a horizontal hyperbola with its center at (1, -2). The transverse axis lies along the x-axis given the negative in front of the y-term.
The transformation from general to standard form involves careful step-by-step manipulation, making sure to observe any changes to symmetry and orientation of the hyperbola as the equation shifts into its standard state.

Equation Manipulation

Equation manipulation involves algebraic transformations and simplifications to meet a desired structural form, such as the standard form of a conic section. It requires strategic operations like expanding, factoring, and adding or subtracting terms while ensuring logical equivalency.
In the exercise, there are several manipulation steps:

• Re-arranging and grouping the x and y terms separately
• Factoring out the coefficients of the leading terms (36 for x and -25 for y)
• Applying completing the square on both sides
• Substituting back into the original equation and simplifying by combining like terms
• Finally, dividing all terms to achieve a standard form with a right-hand side of 1

The simplification and transformations thus result in a tighter, recognizable format. This makes examining the geometrical essence of the hyperbola straightforward, emphasizing shifts like orientation and centering. Each manipulation step serves a purpose, opening pathways towards clarity, necessary for visualizing and working with complex equations like hyperbolas.