Math Problem Statement

Convert the equation to standard form by completing the square on x and y.​ Then, graph the hyperbola. Locate the foci and find the equations of the asymptotes. 9 x squared minus 4 y squared minus 54 x plus 32 y minus 19equals0 Question content area bottom Part 1 The standard form of the equation is    enter your response here. ​(Type an equation. Use integers or fractions for any numbers in the​ expression.) Part 2 Graph the hyperbola. Choose the correct graph below. A. -12 6 -12 4 x y

A coordinate system with a horizontal x-axis labeled from negative 6 to 12 in increments of 1, a vertical y-axis labeled from negative 4 to 12 in increments of 1. The pieces opens up and down. The graph has center in third quadrant and the diagonal lines of this rectangel have magnitude of slopes greater than 1 and the curves reaches these lines as they travel away from the center. B. -6 12 -4 12 x y

A coordinate system with a horizontal x-axis labeled from negative 6 to 12 in increments of 2, a vertical y-axis labeled from negative 4 to 12 in increments of 2. The pieces opens up and to the down. The graph has center in first quadrant and the diagonal lines of this rectangle have magnitude of slopes greater than 1 and the curves reaches these lines as they travel away from the center. C. -12 6 -12 4 x y

A coordinate system with a horizontal x-axis labeled from negative 6 to 12 in increments of 1, a vertical y-axis labeled from negative 4 to 12 in increments of 1. The pieces opens to the left and to the right. The graph has center in third quadrant and the diagonal lines of this rectangle have magnitude of slopes greater than 1 and the curves reaches these lines as they travel away from the center. D. -6 12 -4 12 x y

A coordinate system with a horizontal x-axis labeled from negative 6 to 12 in increments of 1, a vertical y-axis labeled from negative 4 to 12 in increments of 1. The pieces opens to the left and to the right. The graph has center in first quadrant and the diagonal lines of this rectangle have magnitude of slopes greater than 1 and the curves reaches these lines as they travel away from the center. Part 3 The foci​ is/are at the​ point(s)    enter your response here. ​(Type an ordered pair. Type an exact​ answer, using radicals as needed. Use a comma to separate answers as​ needed.) Part 4 The equation of the asymptote with the positive slope is    enter your response here. The equation of the asymptote with the negative slope is    enter your response here. ​(Use integers or fractions for any numbers in the​ equation.) The standard form of the equation is(Type an equation. Use integers or fractions for any numbers in the expression.)Choose the correct graph below.The foci is/are at the point(s)(Type an ordered pair. Type an exact answer, using radicals as needed. Use a comma to separate answers as needed.)The equation of the asymptote with the positive slope is.The equation of the asymptote with the negative slope isThe equation of the asymptote with the negative slope is(Use integers or fractions for any numbers in the equation.)

Solution

Part 1: Convert to Standard Form

The given equation is:

9x2−4y2−54x+32y−19=09x^2 - 4y^2 - 54x + 32y - 19 = 0

  1. Group xx- and yy-terms: 9x2−54x−4y2+32y=199x^2 - 54x - 4y^2 + 32y = 19

  2. Factor out the coefficients of x2x^2 and y2y^2: 9(x2−6x)−4(y2−8y)=199(x^2 - 6x) - 4(y^2 - 8y) = 19

  3. Complete the square:

    • For x2−6xx^2 - 6x, half of −6-6 is −3-3, and (−3)2=9(-3)^2 = 9.
    • For y2−8yy^2 - 8y, half of 88 is 44, and 42=164^2 = 16.

    Add these inside the parentheses and adjust the equation to maintain equality: 9(x2−6x+9)−4(y2−8y+16)=19+9(9)−4(16)9(x^2 - 6x + 9) - 4(y^2 - 8y + 16) = 19 + 9(9) - 4(16)

    Simplify: 9(x−3)2−4(y−4)2=19+81−64=369(x - 3)^2 - 4(y - 4)^2 = 19 + 81 - 64 = 36

  4. Divide through by 36 to normalize: 9(x−3)236−4(y−4)236=1\frac{9(x - 3)^2}{36} - \frac{4(y - 4)^2}{36} = 1

    Simplify: (x−3)24−(y−4)29=1\frac{(x - 3)^2}{4} - \frac{(y - 4)^2}{9} = 1

The standard form is:

(x−3)24−(y−4)29=1\frac{(x - 3)^2}{4} - \frac{(y - 4)^2}{9} = 1

Part 2: Graph the Hyperbola

This hyperbola opens left and right because the xx-term comes first in the equation. The center is (3,4)(3, 4), with vertices at (1,4)(1, 4) and (5,4)(5, 4).

The correct graph is D.

Part 3: Foci

For hyperbolas, the distance from the center to each focus is cc, where: c=a2+b2c = \sqrt{a^2 + b^2}

Here: a2=4,b2=9  ⟹  c=4+9=13a^2 = 4, \quad b^2 = 9 \quad \implies \quad c = \sqrt{4 + 9} = \sqrt{13}

The foci are at: (3−13,4)and(3+13,4)(3 - \sqrt{13}, 4) \quad \text{and} \quad (3 + \sqrt{13}, 4)

Part 4: Asymptotes

The equations of the asymptotes for a hyperbola in standard form (x−h)2a2−(y−k)2b2=1\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 are: y−k=±ba(x−h)y - k = \pm \frac{b}{a}(x - h)

Substitute h=3h = 3, k=4k = 4, a=2a = 2, and b=3b = 3: y−4=±32(x−3)y - 4 = \pm \frac{3}{2}(x - 3)

Simplify: y=32(x−3)+4andy=−32(x−3)+4y = \frac{3}{2}(x - 3) + 4 \quad \text{and} \quad y = -\frac{3}{2}(x - 3) + 4

The equations of the asymptotes are: y=32x−12andy=−32x+172.y = \frac{3}{2}x - \frac{1}{2} \quad \text{and} \quad y = -\frac{3}{2}x + \frac{17}{2}.

Final Results

  1. Standard Form: (x−3)24−(y−4)29=1\frac{(x - 3)^2}{4} - \frac{(y - 4)^2}{9} = 1

  2. Correct Graph: D.

  3. Foci: (3−13,4),(3+13,4)(3 - \sqrt{13}, 4), \quad (3 + \sqrt{13}, 4)

  4. Asymptotes: y=32x−12,y=−32x+172.y = \frac{3}{2}x - \frac{1}{2}, \quad y = -\frac{3}{2}x + \frac{17}{2}.

Do you need detailed steps on graphing or further clarifications?

Related Questions:

  1. How do you determine the orientation of a hyperbola?
  2. What are the key differences between hyperbolas and ellipses?
  3. How do you calculate the eccentricity of a hyperbola?
  4. What is the relationship between the slopes of the asymptotes and the hyperbola's axes?
  5. How do you identify the transverse and conjugate axes?

Tip: Always check the denominator values under x2x^2 and y2y^2 to identify the hyperbola's orientation.

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Math Problem Analysis

Mathematical Concepts

Hyperbolas
Conic Sections
Completing the Square
Graphing Hyperbolas

Formulas

Standard form of a hyperbola: (x-h)^2/a^2 - (y-k)^2/b^2 = 1
Distance to foci: c = sqrt(a^2 + b^2)
Equations of asymptotes: y - k = ±(b/a)(x - h)

Theorems

Properties of Hyperbolas
Relationship between center, vertices, and foci in hyperbolas

Suitable Grade Level

Grades 10-12

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