Question 603025
<pre>
All of them have the form of 5 terms:

Ax² + Cy² + Dx + Ey + F = 0

Get them in this form with 0 on the right.  If there are
fewer than 5 terms, then put 0 terms for the missing ones.
For example if you have

2y² + x - 2 = 0

then consider that as the same as

0x² + 2y² + 1x + 0y - 2 = 0

and then

A=0, C=2, D=1, E=0, and F=-2

[Don't confuse the capital "A" and "C" with the little
"a" and "c" in the standard forms of the ellipses and
hyperbolas.]

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To determine which conic section the graph is of,

If A=C then it is a circle.
If A or C is 0 then it is a parabola.
If A and C have the same sign but not equal it is an ellipse.
If A and C have opposite signs then it is a hyperbola.

1. If A=C then it is a CIRCLE and can be placed in the form

(x-h)² + (y-k)² = r² with center (h,k) and radius r

2. If A and C have the same sign but are not equal, then it

is an ELLIPSE and can either be placed in this form:

{{{(x-h)^2/a^2}}} + {{{(y-k)^2/b^2}}} = 1

which looks like this:

{{{drawing(100,50,-2,2,-1,1,arc(0,0,4,-2),
red(line(0,0,2,0)),green(line(0,0,0,1))

)}}}

or it is an ELLIPSE and can either be placed in this form:

{{{(x-h)^2/b^2}}} + {{{(y-k)^2/a^2}}} = 1

which looks like this:

{{{drawing(50,100,-1,1,-2,2,arc(0,0,2,-4),
red(line(0,0,0,2)),green(line(0,0,1,0))

)}}}

where the center is (h,k), "a" is the length of the red line
and "b" is the length of the green line. "a" is always longer
than "b".  You can tell which of the two forms it is by observing
whether the larger which is a² is under the term (x-h)² or under
the term (y-k)².

The foci, or focal points, are the two points on the red line which
are "c" units from the center.  "c" is calculated by c² = a²-b² for
all ellipses.

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3. If A and C have opposite signs, then it

is a HYPERBOLA and can either be placed in this form:

{{{(x-h)^2/a^2}}} - {{{(y-k)^2/b^2}}} = 1

which looks like this:

{{{drawing(160,120,6,14,7,13,graph(160,120,6,14,7,13,10+sqrt(-4+(x-10)^2)/2),
graph(160,120,6,14,7,13,10-sqrt(-4+(x-10)^2)/2),
green(line(10,10,10,11)),red(line(10,10,12,10)),
line(8,9,12,9),line(12,9,12,11),line(12,11,8,11),line(8,11,8,9),
blue(line(-10,0,20,15), line(-2,16,20,5))

)}}}

or it is a HYPERBOLA and can either be placed in this form:

{{{(y-k)^2/a^2}}} - {{{(x-h)^2/b^2}}} = 1

which looks like this:

{{{drawing(160,120,6,14,7,13,graph(160,120,6,14,7,13,10+sqrt(4+(x-10)^2)/2),
graph(160,120,6,14,7,13,10-sqrt(4+(x-10)^2)/2),
red(line(10,10,10,11)),green(line(10,10,12,10)),
line(8,9,12,9),line(12,9,12,11),line(12,11,8,11),line(8,11,8,9),
blue(line(-10,0,20,15), line(-2,16,20,5))

)}}}

where the center is (h,k), "a" is the length of the red line
and "b" is the length of the green line.

[CAUTION: you can't go
by the length of "a" and "b" in a hyperbola as you can with an
ellipse. Sometimes "a" is larger than "b", sometimes b is larger
than a, and sometimes they are equal.]

You can tell which of the two forms it is by observing that a² is
always under the POSITIVE term and b² is always under the negative
term.

The foci, or focal points, are the two points on the extended red
line which are "c" units from the center.  "c" is calculated by
c² = a²+b² for all hyperbolas.  Notice that there is plus sign,
whereas the formula for c in the ellipse there is a minus sign.

The asymptotes are the extended diagonal of the defining rectangle.

-------------------------------

If A=0, then it can be placed in the form

(x - h)² = 4p(y - k)

and looks like one of these:

{{{drawing(100,100,6,14,7,15, graph(100,100,6,14,7,15,10+(x-10)^2/4),
blue(line(8,11,12,11)),red(line(10,10,10,11)),green(line(10,10,10,9)),
line(5,9,15,9) )}}} OR {{{drawing(100,100,6,14,7,15, graph(100,100,6,14,7,15,10-(x-10)^2/4),
blue(line(8,9,12,9)),red(line(10,10,10,9)),green(line(10,10,10,11)),
line(5,11,15,11) )}}}

The vertex is the point (h,k).  The red and green lines are both
"|p|" units long. The blue line is the latus rectum and it is "|4p|"
units long. The focal point, or focus, is the midpoint of the
latus rectum.  The black line is the directrix.  The parabola opens
upward as in the first graph if p is positive and downward if p is
negative.

-----------------------------

If C=0, then it can be placed in the form

(y - k)² = 4p(x - h)

and looks like one of these:

{{{drawing(100,100,6,14,6,14, graph(100,100,6,14,6,14,10+2sqrt(x-10)),
graph(100,100,6,14,6,14,10-2sqrt(x-10)),

blue(line(11,8,11,12)),red(line(10,10,11,10)),green(line(9,10,10,10)),
line(9,0,9,15) )}}} OR {{{drawing(100,100,6,14,6,14, graph(100,100,6,14,6,14,10+2sqrt(-x+10)),
graph(100,100,6,14,6,14,10-2sqrt(-x+10)),

blue(line(9,8,9,12)),red(line(10,10,9,10)),green(line(11,10,10,10)),
line(11,0,11,15) )}}}

The vertex is the point (h,k).  The red and green lines are both
"|p|" units long. The blue line is the latus rectum and it is "|4p|"
units long. The focal point, or focus, is the midpoint of the
latus rectum.  The black line is the directrix.  The parabola opens
RIGHTward as in the first graph if p is positive and LEFTward if p is
negative.

There are also a lot of graphs of these on this site.  There are also