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Magazine Chapter 12: Problem 17
Find the vertices and foci of the hyperbola. Sketch its graph, showing the asymptotes and the foci. $$ 9 y^{2}-x^{2}-36 y+12 x-36=0 $$
Short Answer
Expert verified
Center: (-6, 2); Vertices: (-6, 0), (-6, 4); Foci: (-6, 2卤2鈭�10); Asymptotes: y = 2卤1/3(x+6).
Step by step solution
01
Simplify the equation
Start by rewriting the given equation:\[ 9y^2 - x^2 - 36y + 12x - 36 = 0 \]Group the \(x\) and \(y\) terms separately:\[ 9(y^2 - 4y) - (x^2 + 12x) = 36 \]
02
Complete the square
To complete the square for the \(y\) terms, \[ y^2 - 4y \] can be rewritten as \( (y-2)^2 - 4 \). Remember to balance it by multiplying the constant by 9: \[ 9(y^2 - 4y + 4 - 4) = 9((y-2)^2 - 4) = 9(y-2)^2 - 36 \].For the \(x\) terms, \[ x^2 + 12x \] can be rewritten as \( (x+6)^2 - 36 \): \[ -(x^2 + 12x + 36 - 36) = -(x+6)^2 + 36 \].
03
Write in standard form
Substitute the completed squares back into the equation:\[ 9(y-2)^2 - (x+6)^2 = 0 \]Rearrange this into the standard form of a hyperbola:\[ \frac{(y-2)^2}{4} - \frac{(x+6)^2}{36} = 1 \]
04
Identify the center, vertices, and foci
The center of the hyperbola is \((h, k) = (-6, 2)\). The standard form \[ \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \] indicates a vertical hyperbola.For vertical hyperbolas:- Vertices: \((h, k\pm a) = (-6, 2\pm 2) = (-6, 0)\) and \((-6, 4)\).- \(c = \sqrt{a^2 + b^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10}\)- Foci: \((h, k\pm c) = (-6, 2\pm 2\sqrt{10})\).
05
Define the asymptotes
For a vertical hyperbola, the equations of the asymptotes are:\[ y = k \pm \frac{a}{b}(x-h) = 2 \pm \frac{2}{6}(x+6) \]This simplifies to:\[ y = 2 \pm \frac{1}{3}(x+6) \].These are the straight lines that the hyperbola approaches but never touches.
06
Sketch the graph
Plot the center at \((-6, 2)\), the vertices at \((-6, 0)\) and \((-6, 4)\), and focus points at \(-6, 2\pm 2\sqrt{10}\).Draw the asymptotes based on the equations \(y = 2 \pm \frac{1}{3}(x+6)\).Draw the hyperbola opening in vertical direction passing through the vertices.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vertices of a Hyperbola
The vertices of a hyperbola are crucial points indicating where the hyperbola intersects the line through its center. When finding these points, you'll follow these general steps:
- Start with the hyperbola's standard equation: \[\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \] for a vertical hyperbola.
- 'a' represents the distance from the center to each vertex. 'a' is always positive.
- Add and subtract 'a' from the y-coordinate of the center to find the vertices.
Foci of a Hyperbola
Foci are special points within the hyperbola structure that play a central role in its definition. They reside along the axis of symmetry, further away from the center than the vertices:
- Utilize the standard form to determine 'c', the distance from the center to each focus.
- For vertical hyperbolas, \(c = \sqrt{a^2 + b^2}\).
- Add and subtract 'c' from the y-coordinate of the center to find the foci.
Asymptotes of a Hyperbola
Asymptotes are diagonal lines that a hyperbola approaches but never actually touches. They are essential for understanding the hyperbola's open-ended nature. Here's how you find them:
- For a vertical hyperbola, use the formula: \(y = k \pm \frac{a}{b}(x-h)\).
- 'a' and 'b' are the same values used in the vertices and foci calculations.
- Substitute the center's \(h\) and \(k\) coordinates into the formula.
Completing the Square
Completing the square is a powerful algebraic method used to transform a quadratic expression into a perfect square trinomial. This forms a crucial step in converting the original equation into standard form:
- For quadratics in \(y\), identify and rework the expression: \(y^2 - 4y\) becomes \((y - 2)^2 - 4\) by completing the square.
- Similarly, adjust the quadratic in \(x\): \(x^2 + 12x\) transforms into \((x + 6)^2 - 36\).
- In both cases, ensure the equation remains balanced after rewriting terms.
