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Difficulty: HardConic Sections

A hyperbola in the standard (x,y)(x, y) coordinate plane is defined by the equation 9x2−16y2−36x−32y−124=09x^2 - 16y^2 - 36x - 32y - 124 = 0. One of the foci of this hyperbola is located at the point (f,−1)(f, -1), where f>0f > 0. What is the value of ff?

Answer: 7

Answer

7
Completing the square transforms the equation into the standard form of a horizontal hyperbola, (x−2)216−(y+1)29=1\frac{(x - 2)^2}{16} - \frac{(y + 1)^2}{9} = 1, which has its center at (2,−1)(2, -1) with a2=16a^2 = 16 and b2=9b^2 = 9. The distance to the foci is c=a2+b2=16+9=5c = \sqrt{a^2 + b^2} = \sqrt{16 + 9} = 5. Since the transverse axis is horizontal, the foci are located at (2±5,−1)(2 \pm 5, -1), which are (−3,−1)(-3, -1) and (7,−1)(7, -1). Given the constraint that f>0f > 0, the positive x-coordinate of the focus is 7.

Step-by-Step Solution

1
Group the terms and prepare to complete the square.
9(x2−4x)−16(y2+2y)=1249(x^2 - 4x) - 16(y^2 + 2y) = 124
Grouping the variables helps isolate the quadratic expressions for completing the square.
2
Complete the square for both the xx and yy terms.
9(x−2)2−16(y+1)2=1449(x - 2)^2 - 16(y + 1)^2 = 144
To complete the square for x2−4xx^2 - 4x, add 4 inside the first parentheses, adding 9×4=369 \times 4 = 36 to the right side. To complete the square for y2+2yy^2 + 2y, add 1 inside the second parentheses, which subtracts 16×1=1616 \times 1 = 16 from the right side because of the leading negative coefficient. This leaves the right side as 124+36−16=144124 + 36 - 16 = 144.
3
Divide both sides of the equation by the constant to find the standard form.
(x−2)216−(y+1)29=1\frac{(x - 2)^2}{16} - \frac{(y + 1)^2}{9} = 1
Dividing by 144 puts the equation in the standard horizontal hyperbola form: (x−h)2a2−(y−k)2b2=1\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1.
4
Find the distance cc from the center to the foci.
c=5c = 5
For a hyperbola, the relationship between the semi-axes and the focal distance is c=a2+b2c = \sqrt{a^2 + b^2}. Substituting a2=16a^2 = 16 and b2=9b^2 = 9 gives c=16+9=5c = \sqrt{16 + 9} = 5.
5
Determine the coordinates of the foci and extract the value of ff.
f=7f = 7
The center of the hyperbola is (h,k)=(2,−1)(h, k) = (2, -1). The foci are located at (h±c,k)=(2±5,−1)(h \pm c, k) = (2 \pm 5, -1), which corresponds to the points (−3,−1)(-3, -1) and (7,−1)(7, -1). Since the problem states f>0f > 0, the target focus must be (7,−1)(7, -1), meaning f=7f = 7.

Key Concept

Converting a general hyperbola equation into standard form to calculate focal points
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Topics

Conic Sections

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