Question 973815
Conics with axes of symmetry parallel to the x-axis and/or y-axis are easy to identify and graph.
 
PARABOLAS have only one variable squared, and you can transform them into something like {{{y=a(x-h)^2+k}}} or {{{x=a(y-h)^2+k}}} with some constants {{{a}}} , {{{h}}} , and {{{k}}} .
{{{x^2-6x=y+3}}}
{{{x^2-6x+9=y+3+9}}}
{{{(x-3)^2=y+12}}}
{{{y=(x-3)^2-12}}} {{{drawing(300,300,-2,8,-14,6,grid(1),
graph(300,300,-2,8,-14,6,(x-3)^2-12)
)}}}
 
ELLIPSES and CIRCLES have equations that can be transformed into something like
{{{(x-h)^2/a^2+(y-k)^2/b^2=1}}}
With {{{a=b=R}}} we get a circle: {{{(x-h)^2/R^2+(y-k)^2/R^2=1}}}<--->{{{(x-h)^2+(y-k)^2=R^2}}} .
If {{{a<>b}}} the circle gets stretched to make it longer in the direction of the variable atop that larger of {{{a^2}}} and {{{b^2}}} .
If you see an {{{x^2}}} and a {{{y^2}}} that would both have a positive coefficient if they were on the same side of the equal sign, suspect an ellipse
(you don't have one of those in your examples).
 
HYPERBOLAS have equations that can be transformed into something like
{{{(x-h)^2/a^2-(y-k)^2/b^2=1}}} or  {{{(y-k)^2/a^2-(x-h)^2/b^2=1}}} .
{{{ (x-2)^2/9 - (y+2)^2/4 = 1}}}<--->{{{(x-2)^2/3^2-(y-(-2))^2/2^2=1}}} is already it that form (no need to transform it much).
Its axes of symmetry are the lines {{{x-2=0}}} and {{{y+2=0}}} .
It is centered at (2,-2).
Your other example takes a little more work to get it into that form,
but you may not need to do the transformation.
If the terms in {{{x^2}}} and {{{y^2}}} have coefficients with opposite sign when they are on the same side of the equal sign, suspect a hyperbola.
{{{3x^2 - y^2 - 7x + 2 = 0 }}}
{{{3(x^2 - 7x/3)+7^2/6^2 - y^2 = -2 }}}
{{{3(x^2-7x/3+(7/6)^2-49/36) - y^2 = -2 }}}
{{{3(x^2-7x/3+(7/6)^2)-3(49/36) - y^2 = -2 }}}
{{{3(x-7/6)^2-49/12 - y^2 = -2 }}}
{{{3(x-7/6)^2 - y^2 = 49/12-2 }}}
{{{3(x-7/6)^2 - y^2 = 25/12 }}}
Now, I divide by {{{25/12}}} (or multiply times {{{12/25}}} , same thing) to get
{{{3(12/25)(x-7/6)^2 - (12/25)y^2 = (12/25)(25/12) }}}
{{{(36/25)(x-7/6)^2 - (12/25)y^2 = 1}}}
{{{(x-7/6)^2/(25/36) - y^2/(25/12) = 1}}}
{{{(x-7/6)^2/(5/6)^2 - y^2/(25/12) = 1}}}
That is really a hyperbola, because teachers usually do not try to trick you.
 
NOTE:
If the teacher wanted to trick you, he/she could give you something that had an {{{x^2}}} and a {{{-y^2}}} , but transformed into something like
{{{(x-2)^2-(y+2)^2=0}}}<-->{{{(y+2)^2=(x-2)^2}}}<--->{{{system(y+2=x-2,"or",y+2=-(x-2))}}}<--->{{{system(y=x-4,"or",y=-x)}}} ,
and that graphs as just two lines that intersect,
but you can call that a degenerate hyperbola,
a hyperbola that degenerated into those two lines.
You could also have something that may have an {{{x^2}}} and a {{{y^2}}} with positive coefficients on the same side of the equal sign,
but when transformed turned out to be a "degenerate ellipse,"
that could be a single point (as for {{{x-2}^2/4+y+2)2/9=0}}} ),
or no point at all (as for {{{x-2}^2/4+y+2)2/9=-1}}} ).