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:

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

or

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

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Geometry of a Hyperbola

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General Form of the Equation

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Geometry of a Hyperbola