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Difficulty: Very hardConic Sections

In the standard (x,y)(x, y) coordinate plane, a hyperbola is defined by the equation 9x2−16y2−54x−64y−127=09x^2 - 16y^2 - 54x - 64y - 127 = 0. What is the shortest distance from the focus of the hyperbola with the larger xx-coordinate to the asymptote with the positive slope?

  1. A
    157\frac{15}{7}
  2. 3Answer
  3. C
    4
  4. D
    5
  5. E
    175\frac{17}{5}

Answer

3
The correct answer is 3. Completing the square for 9x2−16y2−54x−64y−127=09x^2 - 16y^2 - 54x - 64y - 127 = 0 yields the standard form equation (x−3)216−(y+2)29=1\frac{(x-3)^2}{16} - \frac{(y+2)^2}{9} = 1. The focus with the larger xx-coordinate is at (8,−2)(8, -2) and the asymptote with the positive slope is 3x−4y−17=03x - 4y - 17 = 0. Applying the point-to-line distance formula yields a distance of 3.

Step-by-Step Solution

1
Group the xx-terms and yy-terms and move the constant to the right side of the equation.
9(x2−6x)−16(y2+4y)=1279(x^2 - 6x) - 16(y^2 + 4y) = 127
Grouping the terms allows us to complete the square for the xx and yy variables separately.
2
Complete the square for both the xx and yy expressions, adjusting the right side of the equation by adding the weighted constants.
9(x2−6x+9)−16(y2+4y+4)=127+9(9)−16(4)  ⟹  9(x−3)2−16(y+2)2=1449(x^2 - 6x + 9) - 16(y^2 + 4y + 4) = 127 + 9(9) - 16(4) \implies 9(x-3)^2 - 16(y+2)^2 = 144
Completing the square allows us to write the quadratic expressions as perfect squares to put the equation in standard form.
3
Divide both sides of the equation by 144 to obtain the standard form of the hyperbola.
(x−3)216−(y+2)29=1\frac{(x-3)^2}{16} - \frac{(y+2)^2}{9} = 1
The standard form of a horizontal hyperbola is (x−h)2a2−(y−k)2b2=1\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1, which reveals the center (h,k)(h, k) and the semi-axes aa and bb.
4
Identify the key parameters of the hyperbola: center, aa, bb, and calculate the focal distance cc.
Center is (3,−2)(3, -2), a=4a = 4, b=3b = 3, and c=a2+b2=16+9=5c = \sqrt{a^2 + b^2} = \sqrt{16 + 9} = 5.
These parameters are required to find the coordinates of the focus and the equation of the asymptote.
5
Find the coordinates of the focus with the larger xx-coordinate.
Focus is (3+5,−2)=(8,−2)(3 + 5, -2) = (8, -2).
For a horizontal hyperbola, the foci are located at (h±c,k)(h \pm c, k). The focus with the larger xx-coordinate is at (h+c,k)(h+c, k).
6
Determine the equation of the asymptote with the positive slope.
The asymptote equation is y+2=34(x−3)  ⟹  3x−4y−17=0y + 2 = \frac{3}{4}(x - 3) \implies 3x - 4y - 17 = 0.
The asymptotes of a horizontal hyperbola are given by y−k=±ba(x−h)y - k = \pm \frac{b}{a}(x - h). The one with the positive slope uses +ba+\frac{b}{a}.
7
Use the point-to-line distance formula d=∣Ax0+By0+C∣A2+B2d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} to find the distance from the focus (8,−2)(8, -2) to the asymptote line 3x−4y−17=03x - 4y - 17 = 0.
d=∣3(8)−4(−2)−17∣32+(−4)2=∣24+8−17∣25=155=3d = \frac{|3(8) - 4(-2) - 17|}{\sqrt{3^2 + (-4)^2}} = \frac{|24 + 8 - 17|}{\sqrt{25}} = \frac{15}{5} = 3.
Calculating this gives the shortest distance from the focus to the asymptote.

Key Concept

Rewriting the general equation of a hyperbola into standard form by completing the square, identifying its center, foci, and asymptotes, and applying the distance formula from a point to a line.
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Topics

Conic Sections

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