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Magazine Chapter 8: Problem 82
Write the equation in standard form for a hyperbola centered at ( \(h, k\) ). Identify the center and vertices. $$y^{2}+8 y-3 x^{2}+13=0$$
Short Answer
Expert verified
Center: (0, -4); Vertices: (0, -3.73) and (0, -4.27).
Step by step solution
01
Group Terms
Let's start by rearranging the given equation, grouping the like terms together: y terms: \( y^2 + 8y \)x terms: \(-3x^2\).We want to write these grouped terms separately as: \[ y^2 + 8y - 3x^2 = -13 \]
02
Complete the Square for y
To transform the \( y \) terms, we need to complete the square. Look at \( y^2 + 8y \). Half of 8 is 4, and 4 squared is 16.We add and subtract 16 to complete the square:\[ y^2 + 8y + 16 - 16 \]Rewrite as: \[ (y + 4)^2 - 16 \].
03
Rewrite the Equation
Replace \( y^2 + 8y \) with \((y + 4)^2 - 16\) in the equation:\[ (y + 4)^2 - 16 - 3x^2 = -13 \] Simplify this to:\[ (y + 4)^2 - 3x^2 = 3 \].
04
Convert to Standard Form of a Hyperbola
Divide the entire equation by 3 so the right side equals 1:\[ \frac{(y+4)^2}{3} - \frac{3x^2}{3} = \frac{3}{3} \]Simplify to:\[ \frac{(y+4)^2}{3} - x^2 = 1 \]This is the standard form of a hyperbola centered at \((0, -4)\) with the transverse axis parallel to the y-axis.
05
Identify the Center
The equation is in the form \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \), which means the center is \((h, k)\).Here, \(h = 0\) and \(k = -4\), so the center is \((0, -4)\).
06
Find the Vertices
For hyperbolas of form \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \), vertices are located \(a\) units above and below the center along the y-axis because the y-term is positive.Given \(\frac{(y+4)^2}{3}\), \(a^2 = 3\), so \(a = \sqrt{3}\).Vertices are at \((0, -4 + \sqrt{3})\) and \((0, -4 - \sqrt{3})\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Standard Form Equation
The standard form of a hyperbola's equation is essential when dealing with this type of conic section. It's a specific structure that provides clarity on the hyperbola's orientation and dimensions.
- The general standard form for a hyperbola is: \[ \frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1 \] or \[ \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 \].
- The choice between these forms depends on whether the transverse axis is parallel to the y-axis or the x-axis.
- For a hyperbola, the right side of the equation is always 1. This differentiates it from ellipses, which also have similar forms but differ in structure.
- When an equation of a hyperbola is given, it's common to first rearrange and simplify it before dividing by the constant to achieve the standard form.
Vertices of Hyperbola
Vertices of a hyperbola are specific points where the curve intersects its transverse axis. Recognizing them helps to better understand the shape and dimensions of the hyperbola.
- From the standard form \[ \frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1 \], the vertices are found \(a\) units away from the center along the transverse axis.
- In this problem, the equation \[ \frac{(y+4)^2}{3} - x^2 = 1 \] indicates that the center is at \( (0, -4) \) and the transverse axis is vertical since the y-term comes first.
- Since \( a^2 = 3 \), we have \( a = \sqrt{3} \).
- Thus, the vertices are located at \( (0, -4 + \sqrt{3}) \) and \( (0, -4 - \sqrt{3}) \).
Completing the Square
Completing the square is a fundamental technique used to transform quadratic expressions into a perfect square trinomial. This conversion helps in simplifying expressions to either solve or rearrange equations.
- This method is particularly handy when dealing with conic sections like circles, ellipses, and hyperbolas.
- For the quadratic expression \( y^2 + 8y \), the procedure involves taking half of the coefficient of \( y \) (which is 8), squaring it (resulting in 16), and then adding and subtracting this value in the expression: \( y^2 + 8y + 16 - 16 \).
- The expression can then be rewritten as a square: \( (y + 4)^2 - 16 \).
- This makes it possible to easily manipulate the quadratic into a form that relates to the conic section in question, which is crucial for rewriting equations into their standard forms.
