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Magazine Chapter 13: Problem 81
Write the equation \(36 x^{2}-25 y^{2}-72 x-100 y=964\) in standard form to show that it describes a hyperbola.
Short Answer
Expert verified
The equation describes a hyperbola: \(\frac{(x-1)^2}{25} - \frac{(y+2)^2}{36} = 1\).
Step by step solution
01
Rearrange Terms
Start by grouping the terms involving the same variable. Write the equation as the sum of squared terms:\[36x^2 - 72x - 25y^2 - 100y = 964\]
02
Complete the Square for x
Factor out the coefficient of \(x^2\) from the terms involving \(x\). Complete the square for these terms:\[36(x^2 - 2x)\]To complete the square, add and subtract \(1\) (since \((\frac{-2}{2})^2 = 1\)) inside the parentheses:\[36((x^2 - 2x + 1) - 1) = 36((x - 1)^2 - 1)\]So it becomes:\[36(x-1)^2 - 36\]
03
Complete the Square for y
Factor out the coefficient of \(y^2\) from the terms involving \(y\). Complete the square for these terms:\[-25(y^2 + 4y)\]Add and subtract \(4\) (since \((4/2)^2 = 4\)) inside the parentheses:\[-25((y^2 + 4y + 4) - 4) = -25((y + 2)^2 - 4)\]So it becomes:\[-25(y+2)^2 + 100\]
04
Substitute Completed Squares Back into Equation
Replace the squared terms back into the original equation:\[36(x-1)^2 - 36 - 25(y+2)^2 + 100 = 964\]
05
Simplify Equation
Combine the constant terms:\[36(x-1)^2 - 25(y+2)^2 = 964 + 36 - 100\]Simplify the right side:\[36(x-1)^2 - 25(y+2)^2 = 900\]
06
Divide by Constant to Achieve Standard Form
Divide the entire equation by 900 to express the equation in standard form:\[\frac{36(x-1)^2}{900} - \frac{25(y+2)^2}{900} = 1\]This simplifies to:\[\frac{(x-1)^2}{25} - \frac{(y+2)^2}{36} = 1\]
07
Recognize and Confirm Hyperbola Form
The equation \(\frac{(x-1)^2}{25} - \frac{(y+2)^2}{36} = 1\) is in the standard form of a hyperbola, \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), indicating it describes a hyperbola.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Complete the Square
Completing the square is a method used to transform a quadratic expression into a perfect square trinomial. This technique makes it easier to manipulate equations, particularly when converting to the standard form of a hyperbola. To complete the square involves creating a perfect square binomial from a quadratic expression, such as turning \(ax^2 + bx\) into \(a((x-h)^2 + c)\). Here's how you do it:
- First, isolate the quadratic term and the linear term together.
- Factor out the coefficient of the quadratic term (if it's not 1).
- Determine the number to add and subtract within the parentheses to form a perfect square trinomial, using the formula \((\frac{b}{2a})^2\).
- Add and subtract this number inside the expression to maintain equality.
- Rewrite the expression as a square of a binomial and adjust for the constant.
Standard Form
The standard form of a hyperbola's equation is essential because it clearly displays the hyperbola's orientation and the coordinates of the center. It follows the pattern \[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\] or \[\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\],where:
- \((h, k)\) represents the center of the hyperbola.
- \(a^2\) and \(b^2\) are the squares of the transverse and conjugate axes' lengths, respectively.
- The equation determines the orientation: 'horizontal' if \(x\) term is positive and 'vertical' if \(y\) term is positive.
Equation of a Hyperbola
A hyperbola's equation describes a specific type of conic section, defined by its reflective properties and distinct shape. The equation is derived from the geometric definition of a hyperbola as the set of all points where the difference of the distances from two fixed points (the foci) is constant. In hyperbola equations, \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\),
- The minus sign indicates that it's indeed a hyperbola, differing from ellipses where this sign is positive.
- The center \((h, k)\) marks the midpoint between the vertices of the hyperbola.
- The vertices lie on the transverse axis, which is determined by the variable that comes first in the equation (\(x\) or \(y\)).
