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Difficulty: MediumConic Sections

An ellipse in the standard coordinate plane is defined by the equation 9x2+25y2−36x+50y−164=09x^2 + 25y^2 - 36x + 50y - 164 = 0. What is the distance between the two foci of this ellipse?

  1. 8Answer
  2. B
    4
  3. C
    16
  4. D
    2342\sqrt{34}
  5. E
    12

Answer

8
The correct answer is 8. Rearranging the given equation 9x2+25y2−36x+50y−164=09x^2 + 25y^2 - 36x + 50y - 164 = 0 by completing the square gives the standard form (x−2)225+(y+1)29=1\frac{(x-2)^2}{25} + \frac{(y+1)^2}{9} = 1. In this standard horizontal ellipse equation, the semi-major axis squared is a2=25a^2 = 25 and the semi-minor axis squared is b2=9b^2 = 9. The focal distance cc from the center to each focus is found using the relation c2=a2−b2c^2 = a^2 - b^2, which yields c=25−9=4c = \sqrt{25 - 9} = 4. Since the distance between the two foci is 2c2c, the final distance is 2(4)=82(4) = 8.

Step-by-Step Solution

1
Group the xx and yy terms and move the constant to the right-hand side.
(9x2−36x)+(25y2+50y)=164(9x^2 - 36x) + (25y^2 + 50y) = 164
Grouping like variables allows us to factor out coefficients before completing the square.
2
Factor out the leading coefficients of the quadratic terms.
9(x2−4x)+25(y2+2y)=1649(x^2 - 4x) + 25(y^2 + 2y) = 164
Completing the square requires the quadratic terms inside the parentheses to have a coefficient of 1.
3
Complete the square for both variables by adding the square of half the linear coefficients inside the parentheses, and balance the equation by adding the distributed values to the right side.
9(x2−4x+4)+25(y2+2y+1)=164+9(4)+25(1)9(x^2 - 4x + 4) + 25(y^2 + 2y + 1) = 164 + 9(4) + 25(1) which simplifies to 9(x−2)2+25(y+1)2=2259(x-2)^2 + 25(y+1)^2 = 225.
This rewrites the quadratic expressions into perfect square binomials.
4
Divide both sides of the equation by 225 to write the equation in standard form.
(x−2)225+(y+1)29=1\frac{(x-2)^2}{25} + \frac{(y+1)^2}{9} = 1
The standard form of an ellipse equation is equal to 1.
5
Identify the values of a2a^2 and b2b^2 to calculate the distance cc from the center to each focus.
a2=25a^2 = 25 and b2=9b^2 = 9. Using c2=a2−b2c^2 = a^2 - b^2, we get c2=25−9=16c^2 = 25 - 9 = 16, so c=4c = 4.
For a horizontal ellipse, the larger denominator is a2a^2 and the smaller is b2b^2, and the focal distance satisfies c2=a2−b2c^2 = a^2 - b^2.
6
Multiply the focal distance from the center by 2 to find the total distance between the two foci.
Distance=2c=2(4)=8\text{Distance} = 2c = 2(4) = 8.
The distance between the two foci is the length of the segment connecting them, which is centered at (2,−1)(2, -1) and extends cc units in both horizontal directions.

Key Concept

Rewriting an ellipse equation in standard form to determine its key geometric features including its foci.
Estimated Time:1m 30s
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Topics

Conic Sections

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