General Form of the Equation

Key Questions

  • How do I use completing the square to convert the general equation of a hyperbola to standard form?

    I can not remember how to deal with #xy# so I will be demonstrating completing the square without one.

    Consider the equation below

    #9x^2 -16y^2 -36x - 96y - 252 = 0#

    The first thing that we should do is group our #x#s and #y#s

    #9x^2 -16y^2 -36x - 96y - 252 = 0#
    #=> (9x^2 -36x) + (-16y^2 - 96y) - 252 = 0#

    Now, let's make our work easier by factoring out #x^2#'s and #y^2#'s coefficient

    #(9x^2 -36x) + (-16y^2 - 96y) - 252 = 0#

    #=> 9(x^2 - 4x) + -16(y^2 + 6y) - 252 = 0#

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

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

    when a = 1, we have

    #(x + b)^2 = x^2 + 2b + b^2#

    Now, in our equation,

    #9(x^2 - 4x) + -16(y^2 + 6y) - 252 = 0#

    We want #x^2 - 4x# and #y^2 + 6y# to be perfect squares.
    In order to do that, we need to add a 3rd element.

    We know that in #x^2 - 4x#,

    #2b = -4 #
    #b = -2#

    For #x^2 - 4x# to be a perfect square, we need to add #b^2 = 4#

    Meanwhile, for #y^2 + 6y#,

    #2b = 6#
    #b = 3#

    For #y^2 + 6y# to be a perfect square, we need to add #b^2 = 9#

    #9(x^2 - 4x) + -16(y^2 + 6y) - 252 = 0#

    #=> 9(x^2 - 4x + 4) + -16(y^2 + 6y + 9) - 252 = 0#

    We added something in the left-hand side of the equation.
    Since we our dealing with an equality, we need to maintain the
    equality.
    We can do this by adding the same value in the right-hand side of the equation or by subtracting the same value in the left-hand side.

    For this demonstration, I will subtract the same value in the left-hand side

    #=> 9(x^2 - 4x + 4) + -16(y^2 + 6y + 9) - 252 = 0#

    #=> 9(x^2 - 4x + 4) + -16(y^2 + 6y + 9) - 252 - 9(4) - -16(9) = 0#

    Now that we have our perfect square trinomials and our equality remains to be one, we can proceed to simplifying the equation

    #9(x^2 - 4x + 4) + -16(y^2 + 6y + 9) - 252 - 9(4) - -16(9) = 0#

    #=> 9(x - 2)^2 + -16(y + 3)^2 - 252 - 9(4) - -16(9) = 0#
    #=> 9(x - 2)^2 + -16(y + 3)^2 - 252 - 36 - -144 = 0#
    #=> 9(x - 2)^2 + -16(y + 3)^2 - 144 = 0#
    #=> 9(x - 2)^2 + -16(y + 3)^2 = 144#

    #=> (9(x - 2)^2 + -16(y + 3)^2 = 144)/144#

    #=> (9(x - 2)^2)/144 + (-16(y + 3)^2)/144 = 144/144#

    #=> ((x - 2)^2)/16 + (-(y + 3)^2)/9 = 1#

    #=> (x - 2)^2/16 - (y + 3)^2/9 = 1#

    Now we have transformed our general equation to standard form.
    Take note that we can do the same trick to transform other conic sections into their standard form.

    seph · 1 · Oct 25 2014
  • How do you determine the slant asymptotes of a hyperbola?

    Let us find the slant asymptotes of a hyperbola of the form

    #x^2/a^2-y^2/b^2=1#.

    By subtracting #x^2/a^2#,

    #=>-y^2/b^2=-x^2/a^2+1#

    by multiplying by #-b^2#,

    #=> y^2=b^2/a^2 x^2-b^2#

    by taking the square-root,

    #=> y=pm sqrt{b^2/a^2 x^2-b^2}#

    For large #x#, #-b^2# in the square-root is negligible,

    #y=pm sqrt{b^2/a^2 x^2-b^2}approx pm sqrt{b^2/a^2 x^2}=pm b/a x#

    Hence, the slant asymptotes are #y=pm b/a x#.

    I hope that this was helpful.

    Wataru · 2 · Nov 10 2014
  • How do you graph the equation of a hyperbola written in the general form?

    Well a hyperbola will look something like this:

    enter image source here

    So in general terms we can say that an hyperbolic equation will be in the terms

    enter image source here

    or

    enter image source here

    So you can see also from this that it may be in a form like this:

    #(x/a + y/b)(x/a - y/b) = 1#

    or

    #(x/a + y/b)(x/a - y/b) = -1#

    Sidharth · 1 · Feb 15 2015
  • What is the general form of a hyperbola?

    The general form of a hyperbola is

    #Ax^2 + Bx + Cy^2 + Dy + E = 0#

    where either A or C is negative (but never both)

    seph · · Oct 15 2014

Questions

Geometry of a Hyperbola