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Homework Help: Hyperbola and Ellipse's definition

  1. Jun 24, 2011 #1
    1. The problem statement, all variables and given/known data

    An ellipse is a set of all points from two points called a focus (together a foci) has the sum of 2a

    |d1 + d2| = 2a

    A hyperbola is the same except it is difference.

    Now my question is, just who came up with these definitions that it must equal to 2a?? Because if I don't remember that it is 2a, then I will never be able to understand the formulas.



    3. The attempt at a solution

    attempedt at daydreaming an answer, but failed.
     
  2. jcsd
  3. Jun 24, 2011 #2

    HallsofIvy

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    An ellipse is a figure such that the sum of the distances from a point on the graph to the two foci is a constant. A hyperbola is a figure such that the differences of the two distance from a point to the two foci is a constant. The fact that that constant is "2a" comes only after you write the ellipse as
    [tex]\frac{x^2}{a^2}+ \frac{y^2}{b^2}= 1[/tex]
    or
    [tex]\frac{x^2}{b^2}+ \frac{y^2}{a^2}= 1[/tex]
    with a the length of the "major semi-axis" so that 2a is the distance between the two vertices along the longer axis.
    To see that, set up and ellipse with major axis along the x-axis, center at (0, 0). Then the two foci are at (f, 0) and (-f, 0), the vertices along the line between foci are at (a, 0) and (-a, 0). The distance from vertex (a, 0) to the focus (f, 0) is a- f. The distance from that vertex to the other vertex is a from (a, 0) to the center, (0, 0), is a, of course, then the distance from (0, 0) to the other focus, (-f, 0), is f for a total distance from (a, 0) to (-f, 0) of a+ f. The total distance from (a, 0) to the two foci is (a- f)+(a+ f)= 2a. Given that that sum of distances is a constant, that constant must be 2a.

    Similarly the constant for the hyperbola is 2a only after you write the hyperbola as
    [tex]\frac{x^2}{a^2}- \frac{y^2}{b^2}= 1[/tex]
    or
    [tex]\frac{y^2}{a^2}- \frac{x^2}{b^2}= 1[/tex]
    with a the distance from the center of the hyperbola to s vertex.

    Of course, the difference between an ellipse and a hyperbola is that the foci of an ellipse are inside the vertices while for a hyperbola, they are outside.

    Again, set up the hyperbola so its center is at (0, 0), its vertices are at (-a, 0) and (a, 0), its foci at (-f,0) and (f, 0). The distance from (a, 0) to (f, 0) if f- a (the foci are now outside so f> a). The distance from (a, 0) to (-f, 0) is the distance from (a, 0) to (0, 0), a, plus the distance from (0, 0) to (-f, 0), which is f. The distance from (a, 0) to (-f, 0) is a+ f. The difference of those two distances is (a+ f)- (f- a)= 2a.
     
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