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Difficulty: HardConic Sections

A hyperbola in the standard coordinate plane is represented by the equation 9x2−16y2−36x−32y−124=09x^2 - 16y^2 - 36x - 32y - 124 = 0. Which of the following equations represents one of the asymptotes of this hyperbola?

  1. A
    y=34x−12y = \frac{3}{4}x - \frac{1}{2}
  2. B
    y=43x−113y = \frac{4}{3}x - \frac{11}{3}
  3. y=34x−52y = \frac{3}{4}x - \frac{5}{2}Answer
  4. D
    y=34x+52y = \frac{3}{4}x + \frac{5}{2}
  5. E
    y=43x−53y = \frac{4}{3}x - \frac{5}{3}

Answer

y=34x−52y = \frac{3}{4}x - \frac{5}{2}
The correct equation is found by rewriting the hyperbola equation in standard form (x−2)216−(y+1)29=1\frac{(x - 2)^2}{16} - \frac{(y + 1)^2}{9} = 1 through completing the square. This indicates a horizontal hyperbola centered at (2,−1)(2, -1) with a=4a = 4 and b=3b = 3. The asymptotes are given by y−k=±ba(x−h)y - k = \pm \frac{b}{a}(x - h). Substituting the values gives y+1=±34(x−2)y + 1 = \pm \frac{3}{4}(x - 2). Simplifying the positive case results in the correct equation.

Step-by-Step Solution

1
Group the xx-terms and yy-terms and factor out the coefficients of the squared terms.
9(x2−4x)−16(y2+2y)=1249(x^2 - 4x) - 16(y^2 + 2y) = 124
Grouping prepares the algebraic expression for completing the square for both variables.
2
Complete the square for both the xx and yy expressions by adding the balanced constants to the right side.
9(x−2)2−16(y+1)2=1449(x - 2)^2 - 16(y + 1)^2 = 144
Adding 9(4)=369(4) = 36 and subtracting 16(1)=1616(1) = 16 to the right side balances the equation after completing the square.
3
Divide both sides by 144144 to express the equation in the standard form of a hyperbola.
(x−2)216−(y+1)29=1\frac{(x - 2)^2}{16} - \frac{(y + 1)^2}{9} = 1
The standard form equation (x−h)2a2−(y−k)2b2=1\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 reveals the center (h,k)(h, k) and the semi-axes values aa and bb.
4
Identify the key parameters of the hyperbola and state the general formula for its asymptotes.
Center (h,k)=(2,−1)(h, k) = (2, -1), a=4a = 4, b=3b = 3. The asymptotes are given by y−k=±ba(x−h)y - k = \pm \frac{b}{a}(x - h), which simplifies to y+1=±34(x−2)y + 1 = \pm \frac{3}{4}(x - 2).
For a horizontal hyperbola, the rise-over-run slope of the asymptotes is governed by the ratio ba\frac{b}{a}.
5
Simplify the positive slope case to find the matching slope-intercept equation.
y=34x−52y = \frac{3}{4}x - \frac{5}{2}
Distributing the slope gives y+1=34x−32y + 1 = \frac{3}{4}x - \frac{3}{2}, and subtracting 11 from both sides yields the final equation.

Key Concept

Rewriting a hyperbola equation using completing the square to find its asymptotes.

Alternative Method

Instead of completing the square entirely, find the center of the hyperbola by taking partial derivatives. The derivative with respect to xx is 18x−36=0  ⟹  x=218x - 36 = 0 \implies x = 2. The derivative with respect to yy is −32y−32=0  ⟹  y=−1-32y - 32 = 0 \implies y = -1. Thus, the center is (2,−1)(2, -1). The slope of the asymptotes can be found from the ratio of the square roots of the coefficients of the quadratic terms: m=±916=±34m = \pm \sqrt{\frac{9}{16}} = \pm \frac{3}{4}. Using the point-slope form with the center (2,−1)(2, -1) and slope 34\frac{3}{4} gives y+1=34(x−2)y + 1 = \frac{3}{4}(x - 2), which simplifies directly to y=34x−52y = \frac{3}{4}x - \frac{5}{2}.
Estimated Time:2m 30s
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Topics

Conic Sections

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