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Difficulty: MediumConic Sections

An ellipse in the standard (x,y)(x, y) coordinate plane is defined by the equation (x−4)2169+(y+3)2144=1\frac{(x-4)^2}{169} + \frac{(y+3)^2}{144} = 1. What is the distance between the two foci of this ellipse?

Answer: 10

Answer

The distance between the two foci of the ellipse is 10.
By comparing the given equation to the standard form of an ellipse, we find a2=169a^2 = 169 and b2=144b^2 = 144. The distance from the center to each focus, cc, is given by c=a2−b2=169−144=25=5c = \sqrt{a^2 - b^2} = \sqrt{169 - 144} = \sqrt{25} = 5. The total distance between the two foci is 2c=2(5)=102c = 2(5) = 10.

Step-by-Step Solution

1
Identify a2a^2 and b2b^2 from the given equation of the ellipse.
a2=169a^2 = 169 and b2=144b^2 = 144
The standard equation of a horizontal ellipse is (x−h)2a2+(y−k)2b2=1\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1, where a2a^2 is the larger denominator.
2
Calculate the value of cc, the distance from the center to a focus.
c=5c = 5
For an ellipse, the focal distance cc is related to the semi-major axis aa and semi-minor axis bb by the equation c2=a2−b2c^2 = a^2 - b^2.
3
Calculate the distance between the two foci, which is 2c2c.
10
The distance between the two foci of an ellipse is twice the distance from the center to each focus (2c2c).

Key Concept

Focal distance of an ellipse
Estimated Time:1m 15s
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Topics

Conic Sections

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