Question 943929
Parabolas are single curves that can open up or down.
There is no "or" with hyperbolas;
a hyperbola can open up AND down,
or it could open left AND right,
or it could open southeast and northwest,
or .....
Hyperbolas, like most people, have 2 arms (two separate curves, or "branches").
Those arms are flirty curves that approach the center,
only to turn around before getting to that center,
and then go away along a symmetric path.
A popular kind of hyperbola opens to the left and right,
and (with its center) looks like parentheses drawn the wrong way around a dot/point:
{{{")" * "("}}}
Another popular shape is with the openings towards top and bottom:
{{{graph(300,300,1,11,1,11,6+sqrt(1+(x-6)^2),6-sqrt(1+(x-6)^2),6+sqrt(0.0025-(x-6)^2))}}} .
There are tilted hyperbolas too, like
{{{graph(300,300,1,11,1,11,6+1/(x-6),x^2+10,6+sqrt(0.0016-(x-6)^2))}}} , but those make calculations too complicated,
so you will (hopefully) not see that.
 
The curve represented by your equation is a hyperbola.
We know that because the terms in {{{x^2}}} and {{{y^2}}} have coefficients (8 and -10) with opposite signs.
You may be able to figure out which ways the hyperbola opens through some formula given in class,
but I have to work up what that hyperbola looks like to understand it.
(Besides, I cannot memorize formulas, so I resort to understanding and thinking through the problem).
To figure out what your hyperbola looks like, you can "complete the squares":
{{{8x^2 - 10y^2 - 48x - 100y - 98 = 0}}}
{{{8x^2 - 48x - 10y^2 - 100y = 98}}}
Now I am going to add some terms to both sides of the equal sign to get an equivalent equation
{{{8x^2 - 48x + 72 - 10y^2 - 100y - 250 = 98 + 72 - 250}}}
There is a method to mu madnes, as you will soon see:
{{{8(x^2-6x+9) - 10(y^2-10y-25) = -80}}}
{{{8(x-3)^2 - 10(y-5)^2 = -80}}}
Do you see what I meant by "completing the squares"?
I was building up {{{8(x-3)^2}}} from {{{8x^2 - 48x}}} by adding {{{72}}} ,
and I was building up {{{- 10(y-5)^2}}} from {{{- 10y^2 - 100y}}} by adding {{{ - 250}}} .
From {{{8(x-3)^2 - 10(y-5)^2 = -80}}} ,
I divide both sides of the equal sign by {{{(-80)}}} to get a {{{1}}} on the right side:
{{{8(x-3)^2/(-80) - 10(y-5)^2/(-80) = -80/(-80)}}}
{{{-(x-3)^2/10 + (y-5)^2/8 = 1}}}
We can rearrange the equation to
{{{(y-5)^2/8-(x-3)^2/10 = 1}}} if we want.
The last equation looks more like the textbook
{{{(y-k)^2/a^2-(x-h)^2/b^2 = 1}}} for a hyperbola opening up and down.
(I just checked).
Without studying the book, you can realize that the curve represented by the equation
{{{(y-5)^2/8-(x-3)^2/10 = 1}}}
1) is symmetrical with respect to the lines
{{{x-3=0}}}<--->{{{x=3}}} and
{{{y-5=0}}}<--->{{{y=5}}} ,
because that equation does not distinguish between {{{x-3=1}}} and {{{x-3=-1}}} ,
or between {{{x-3=2}}} and {{{x-3=-2}}} ,
or between {{{y-5=1}}} and {{{y-5=-1}}} ,
or between {{{y-5=2}}} and {{{y-5=-2}}} ;
2) you also realize that as {{{(x-3)^2>=0}}}<--->{{{(y-5)^2/8=1}}}<--->
{{{(y-5)^2>=8}}}<--->{{{system(y-5>=sqrt(8),"or",y-5<=-sqrt(8))}}}<--->{{{system(y>=5+sqrt(8),"or",y<=5-sqrt(8))}}} .
So the hyperbola opens up and down,
with no points between {{{5-sqrt(8)}}} and {{{5+sqrt(8)}}} .
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