# General Form of the Equation

## Key Questions

• I can not remember how to deal with $x y$ so I will be demonstrating completing the square without one.

Consider the equation below

$9 {x}^{2} - 16 {y}^{2} - 36 x - 96 y - 252 = 0$

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

$9 {x}^{2} - 16 {y}^{2} - 36 x - 96 y - 252 = 0$
$\implies \left(9 {x}^{2} - 36 x\right) + \left(- 16 {y}^{2} - 96 y\right) - 252 = 0$

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

$\left(9 {x}^{2} - 36 x\right) + \left(- 16 {y}^{2} - 96 y\right) - 252 = 0$

$\implies 9 \left({x}^{2} - 4 x\right) + - 16 \left({y}^{2} + 6 y\right) - 252 = 0$

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

${\left(a x + b\right)}^{2} = {a}^{2} {x}^{2} + 2 a b + {b}^{2}$

when a = 1, we have

${\left(x + b\right)}^{2} = {x}^{2} + 2 b + {b}^{2}$

Now, in our equation,

$9 \left({x}^{2} - 4 x\right) + - 16 \left({y}^{2} + 6 y\right) - 252 = 0$

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

We know that in ${x}^{2} - 4 x$,

$2 b = - 4$
$b = - 2$

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

Meanwhile, for ${y}^{2} + 6 y$,

$2 b = 6$
$b = 3$

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

$9 \left({x}^{2} - 4 x\right) + - 16 \left({y}^{2} + 6 y\right) - 252 = 0$

$\implies 9 \left({x}^{2} - 4 x + 4\right) + - 16 \left({y}^{2} + 6 y + 9\right) - 252 = 0$

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

$\implies 9 \left({x}^{2} - 4 x + 4\right) + - 16 \left({y}^{2} + 6 y + 9\right) - 252 = 0$

$\implies 9 \left({x}^{2} - 4 x + 4\right) + - 16 \left({y}^{2} + 6 y + 9\right) - 252 - 9 \left(4\right) - - 16 \left(9\right) = 0$

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

$9 \left({x}^{2} - 4 x + 4\right) + - 16 \left({y}^{2} + 6 y + 9\right) - 252 - 9 \left(4\right) - - 16 \left(9\right) = 0$

$\implies 9 {\left(x - 2\right)}^{2} + - 16 {\left(y + 3\right)}^{2} - 252 - 9 \left(4\right) - - 16 \left(9\right) = 0$
$\implies 9 {\left(x - 2\right)}^{2} + - 16 {\left(y + 3\right)}^{2} - 252 - 36 - - 144 = 0$
$\implies 9 {\left(x - 2\right)}^{2} + - 16 {\left(y + 3\right)}^{2} - 144 = 0$
$\implies 9 {\left(x - 2\right)}^{2} + - 16 {\left(y + 3\right)}^{2} = 144$

$\implies \frac{9 {\left(x - 2\right)}^{2} + - 16 {\left(y + 3\right)}^{2} = 144}{144}$

$\implies \frac{9 {\left(x - 2\right)}^{2}}{144} + \frac{- 16 {\left(y + 3\right)}^{2}}{144} = \frac{144}{144}$

$\implies \frac{{\left(x - 2\right)}^{2}}{16} + \frac{- {\left(y + 3\right)}^{2}}{9} = 1$

$\implies {\left(x - 2\right)}^{2} / 16 - {\left(y + 3\right)}^{2} / 9 = 1$

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.

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

${x}^{2} / {a}^{2} - {y}^{2} / {b}^{2} = 1$.

By subtracting ${x}^{2} / {a}^{2}$,

$\implies - {y}^{2} / {b}^{2} = - {x}^{2} / {a}^{2} + 1$

by multiplying by $- {b}^{2}$,

$\implies {y}^{2} = {b}^{2} / {a}^{2} {x}^{2} - {b}^{2}$

by taking the square-root,

$\implies y = \pm \sqrt{{b}^{2} / {a}^{2} {x}^{2} - {b}^{2}}$

For large $x$, $- {b}^{2}$ in the square-root is negligible,

$y = \pm \sqrt{{b}^{2} / {a}^{2} {x}^{2} - {b}^{2}} \approx \pm \sqrt{{b}^{2} / {a}^{2} {x}^{2}} = \pm \frac{b}{a} x$

Hence, the slant asymptotes are $y = \pm \frac{b}{a} x$.

I hope that this was helpful.

• 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:

$\left(\frac{x}{a} + \frac{y}{b}\right) \left(\frac{x}{a} - \frac{y}{b}\right) = 1$

or

$\left(\frac{x}{a} + \frac{y}{b}\right) \left(\frac{x}{a} - \frac{y}{b}\right) = - 1$

• The general form of a hyperbola is

$A {x}^{2} + B x + C {y}^{2} + D y + E = 0$

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