Ellipses and Hyperbolae   Draw an Ellipse With a String and Two Fixed Points Geometrically an ellipse is defined as follows:  Let P and Q be fixed points in the plane and let k be a positive real number.  Then a point R is in the ellipse if the sum of the distances from P to R and from Q to R is k         PR + QR  =  k For an interactive view of this concept click here     The Standard Form of an Ellipse Centered at The Origin Recall that the equation of a circle centered at the origin has equation         x2 +  y2  =  r2 where r is the radius. Dividing by r2 we have        x2             y2                    +               =   1                         r2             r2 for an ellipse there are two radii, so that we can expect that the denominators should be different.  Hence we have the standard form of an ellipse centered at the origin:         x2             y2                    +               =   1                         a2             b2   The points (a,0), (-a,0), (0,b), and (0,-b) are called the vertices of the ellipse. Note:  Although we have written the a below the x and the b below the y, it is customary to let a be the larger of the two and b be the smaller.   Example Graph the ellipse x2               y2                     +                 =   1           25               16Solution First identify a and b, remembering to take square roots.         a  =  5        b  =  4 Next plot the vertices          (-5,0), (5,0), (0,4), and (0,-4) Finally, connect the dots.  The graph is shown to the right.   Example Graph the ellipse         4x2 + y2  =  16  Solution This is not in standard form since the right hand side is not 1.  To rectify this, we just divide by 36 to get           4x2              y2                     +                 =   1           16               16 or since 9/36 = 1/4, we get x2               y2                     +                 =   1            4               16 Now we can sketch the graph.  Notice that 16 is larger than 4 so we let a be the square root of 16 and b be the square root of 4.  We have         a  =  4        and          b  =  2 We plot the vertices          (0,4), (0,-4), (2,0), and (-2,0)and connect the vertices with a conic as shown to the right.  Example Graph the ellipse          4x2 + 9y2 = 4 Solution:   First divide by 4         x2 + 9/4y2 = 1Since the 9/4 is not in the denominator, we need to use the following fact about division of numbers         9                  1                    =                          4                4/9This comes from looking at the right hand side and and noticing that it is just a division of the fraction 1/1 by 4/9, which becomes a multiplication of 1/1 by 9/4.        x2             y2                    +                =   1                         1             4/9 so that          a = 1     and     b = 2/3   Application:  Astronomy Suppose that an asteroid orbits the sun (in an elliptical orbit).  And suppose that the longest opposite ends of the orbit are 800 million miles apart and that the shortest opposite ends are 200 million miles apart.  Give an equation for the orbit of the asteroid. Solution:   We have          2a  =  800 million miles  and          2b  =  200 million miles thus         a  =  400 million miles     and     b  =  100 million miles so that                  x2                                          y2                                          +                                   =   1                                                                      (400,000,000)2              (100,000,000)2           The Hyperbola Recall the equation of the ellipse:         x2             y2                    +               =   1                         a2             b2 If instead of a "+" we have a "-", we end up with a different conic called the hyperbola.            x2             y2                    -               =   1                         a2             b2  Example    Sketch the graph of        x2             y2                    -               =   1                         4              9  Solution Check for intercepts: If x = 0 then               y2             -         =  1 which has no solution               9 If y = 0 then             x2                     =  1            4          x2  =  4  so that          x  =  2     or     x  =  -2 If instead of the 1, we have a 0 then         x2             y2                    -               =   0                        4             9 so that         x2             y2                    =                                    4              9 hence         y = (3/2)x     or     y = -(3/2)x These two lines are called the asymptotes of the hyperbola and are found by         y   = b/a   To plot the hyperbola with equation           x2             y2                    -               =   1                         a2             b2  we follow these steps:   Plot the intercepts          (a,0),   (-a,0)  and the points          (0,b), (0,-b)  as with the ellipse. Draw a rectangle containing these four points. Draw the lines that contain the diagonals of the rectangle (these are the asymptotes) Draw the hyperbola that with the vertices          (a,0) and (-a,0)  that has the drawn asymptotes.   Following these steps, to sketch the graph of          x2             y2                    -               =   1                         4             16 We have         a  =  2        and        b  =  4 The vertices are at          (2,0)    and    (-2,0) and the helper points are at         (0,4)    and    (0,-4)Plot these points.  Then draw the rectangle with vertices (2,4), (-2,4), (-2,-4), and (2,-4).  Next draw the two lines through the opposite vertices of the rectangle.  These are the asymptotes.  Finally draw the hyperbola.  On the right this process is shown.            Example: Sketch the graph of         9x2  -  4y2   =   16  First we have to divide by 16 to get           x2               y2                        -                =   1                         16/9             4 We see that          a = 4/3    and    b = 2     Note: If the equation is        y2             x2                    -               =   1                         a2            b2 we follow the same procedure except that (0,a) and (0,-a) are the vertices instead of (b,0) and (-b,0).  Example:          y2                                 -   x2    =   1                         16            Here          b = 1     and     a = 4 The intercepts are (0,4) and (0,-4) and the other two convenient points that make up the fundamental rectangle are (1,0) and (0,1).  The graph is shown below Back to the Intermediate Algebra (Math 154) Home Page e-mail Questions and Suggestions