Standard Form of the Equation

Key Questions

  • How do I know whether the major axis of an ellipse is horizontal or vertical?

    Locate the larger of the 2 denominators. Then look at that numerator.

    If it has an #x# then that is the major axis. Horizontal
    If it has an #y# then that is the major axis. Vertical

    #x^2/4+y^2/16=1#

    The major axis is #y#. Vertical

    #x^2/25+y^2/16=1#

    The major axis is #x#. Horizontal

    AJ Speller · 11 · Sep 28 2014
  • What is meant by an ellipse in standard form?

    The standard form of the ellipse, centered in the point #C(x_C,y_C)# and with the semi-axes #a#, horizontal and #b#, vertical is:

    #(x-x_C)^2/a^2+(y-y_C)^2/b^2=1#.

    Massimiliano · 2 · Mar 2 2015
  • What do #a# and #b# represent in the standard form of the equation for an ellipse?

    For ellipses, #a >= b# (when #a = b#, we have a circle)

    #a# represents half the length of the major axis while #b# represents half the length of the minor axis.

    This means that the endpoints of the ellipse's major axis are #a# units (horizontally or vertically) from the center #(h, k)# while the endpoints of the ellipse's minor axis are #b# units (vertically or horizontally)) from the center.

    The ellipse's foci can also be obtained from #a# and #b#.
    An ellipse's foci are #f# units (along the major axis) from the ellipse's center

    where #f^2 = a^2 - b^2#

    Example 1:

    #x^2/9 + y^2/25 = 1#

    #a = 5#
    #b = 3#

    #(h, k) = (0, 0)#

    Since #a# is under #y#, the major axis is vertical.

    So the endpoints of the major axis are #(0, 5)# and #(0, -5)#

    while the endpoints of the minor axis are #(3, 0)# and #(-3, 0)#

    the distance of the ellipse's foci from the center is

    #f^2 = a^2 - b^2#

    #=> f^2 = 25 - 9#
    #=> f^2 = 16#
    #=> f = 4#

    Therefore, the ellipse's foci are at #(0, 4)# and #(0, -4)#

    Example 2:

    #x^2/289 + y^2/225 = 1#

    #x^2/17^2 + y^2/15^2 = 1#

    #=> a = 17, b = 15#

    The center #(h, k)# is still at (0, 0).
    Since #a# is under #x# this time, the major axis is horizontal.

    The endpoints of the ellipse's major axis are at #(17, 0)# and #(-17, 0)#.

    The endpoints of the ellipse's minor axis are at #(0, 15)# and #(0, -15)#

    The distance of any focus from the center is

    #f^2 = a^2 - b^2#
    #=> f^2 = 289 - 225#
    #=> f^2 = 64#
    #=> f = 8#

    Hence, the ellipse's foci are at #(8, 0)# and #(-8, 0)#

    seph · 1 · Oct 16 2014
  • How do I use completing the square to rewrite the equation of an ellipse in standard form?

    To demonstrate, let's try transforming the equation below into standard form.

    #4x^2 + 9y^2 + 72y - 24x + 144 = 0#

    The first thing to do is group #x#s and #y#s together

    #(4x^2 - 24x) + (9y^2 + 72y) + 144#

    Factor out #x^2#'s and #y^2#'s coefficient

    #4(x^2 - 6x) + 9(y^2 + 8y) + 144 = 0#

    Before we continue, let's recall what happens when a binomial is squared

    #(ax + b)^2 = a^2x^2 + 2abx + b^2#

    For our problem, we want #x^2 - 6x# and #y^2 + 8y# to be perfect squares

    For #x#, we know that

    #a^2 = 1#
    #=> a = 1, a = -1#

    #2ab = -6#
    #2(1)b = -6#
    #2b = -6#
    #b = -3#
    #=> b^2 = 9#

    Note that substituting #a = -1# will result to the same #b^2#

    Hence, to complete the square, we need to add #9#

    Meanwhile, for #y#

    #a^2 = 1#
    #=> a = 1, a = -1#

    #2ab = 8#
    #2(1)b = 8#
    #2b = 8#
    #b = 4#
    #=> b = 16#

    Now, let's add our #b^2#s into the equation.

    #4(x^2 - 6x) + 9(y^2 + 8y) + 144 = 0#

    #4(x^2 - 6x + 9) + 9(y^2 + 8y + 16) + 144 = 0#

    We added something into the left-hand side, to retain the "equality", we need to add the same value to the right-hand side (or substract the value again from the left-hand side)

    #4(x^2 - 6x + 9) + 9(y^2 + 8y + 16) + 144 = 0 + 4(9) + 9(16)#
    #4(x^2 - 6x + 9) + 9(y^2 + 8y + 16) + 144 = 180#

    #=> 4(x - 3)^2 + 9(y + 4)^2 + 144 = 180#
    #=> 4(x - 3)^2 + 9(y + 4)^2 = 180 - 144#
    #=> 4(x - 3)^2 + 9(y + 4)^2 = 36#

    #=> (4(x - 3)^2 + 9(y + 4)^2 = 36)/36#

    #=> (x - 3)^2/9 + (y + 4)^2/4 = 1#

    seph · · Oct 19 2014

Questions

Geometry of an Ellipse