Question

he equation 3x^2+2y^2=0 represents a
a. circle
b. parabola

c. pair of lines
d.none of the above

Answer 1

circle

Answer 2

but i think that the coefficients of x and y must be 1 when its a circle

Answer 3

no actually it might be just one line. Try putting x on the other side and try to get y by itself. Then it looks more like a line

Answer 4

the question is this correct?
\[3x^{2}+2y^{2}=0\]

Answer 5

yes

Answer 6

ellipse with radius zero?

Answer 7

lets see, on a graph when you have the coefficients x and y, they what?

Answer 8

if there is no radius then it isn't an elipse

Answer 9

That's my point.

Answer 10

they intersect, od then it can be a circle.

Answer 11

then is it a cirlcle?

Answer 12

can't*

Answer 13

They're squared, so then it COULD be a parabula,

Answer 14

OR they COULD be a pair on lines.

Answer 15

i believe you have to solve it to find out. :)

Answer 16

Solving for y:
\[y= \pm \sqrt{\frac{-3x^2}{2}}\]

Answer 17

It's a graph of the origin.

Answer 18

you dont have to un sqaire it. you know. you just multiply it out.
\[3x^{2}+2y^{2}=0\]

Answer 19

Multiply what? It's already in simplest form...
("That's my point." <-- get this joke yet?)

Answer 20

(Solve for x/y then substitute)

Answer 21

Substitution would be redundant, except to verify the solution. Looking at the formula for y, it is clear that there is only one solution (the ellipse with radius = 0).

Answer 22

i already know the answer, im helping him find it, an no its not. its a quadratic equation. so how can it be the ellipse with the radius=0?

Answer 23

There is no algebra necessary here, only reasoning.

Answer 24

1.std eqn. for circle is x^2+y^2=a^2
2.if h^2-ab=or>0 then it represents a pol
thats why im confused

Answer 25

What would the graph of an ellipse with radius = 0 look like?

Answer 26

The coefficients 3 and 2 are irrelevant; it is two square numbers adding up to zero - what do those two numbers have to be then?

Answer 27

ohh, but Cliff, this requires algebra, :) ik my math.

Answer 28

Yes, you could use algebra to rearrange the equation like I did when I solved for y if that helps you see the solution easier, but it's not necessary.
Experience with general quadratic equations and conic sections provide the standard form of an equation for an ellipse: ax^2 + by^2 = c where a, b, and c are non-negative.
c affects the size of the ellipse. If c=0 then the ellipse has no size.

Answer 29

y=sqrt(-1.5x^2) This is a line

Answer 30

pair?

Answer 31

@edwardo24 what?

Answer 32

@edwardo24 you mean: \[y=\sqrt{-1.5x^{2} }\]

Answer 33

yes that is what i mean

Answer 34

(That's also the exact same equation I provided earlier, except you forgot the plus/minus)

Answer 35

And that is *not* the equation for a line (unless you want to call it a line of zero length).

Answer 36

its not a line of 0 length

Answer 37

Why not?

Answer 38

If it is a line, then you should be able to put the equation in the form y=mx+b and tell me what the slope is.

Answer 39

no, thats not the one you gave earlyer. the one you gave was this:
\[y=\pm\sqrt{\frac{-3x^{2} } { 2 }}\]

Answer 40

Has anyone even tried putting values of x into that function to see what happens?

Answer 41

we don't know what x is, remember?

Answer 42

I don't see why this is so difficult to understand. It is an ellipse of zero size. The graph of that equation is the origin.

Answer 43

no its not.

Answer 44

It's a single equation with two variables, that makes one of them a free parameter and you can choose values arbitrarily.

Answer 45

Are you kidding me? Then tell me what it looks like then.

Answer 46

look, this is annoying me, its a quadratic equation:
\[x^{2}+y^{2}=0\] just with coefficients 3 and 2.

Answer 47

Go on..

Answer 48

it the perabula. not a freaken circle. if there are both \[x^{2}\] and \[y{2}\] that equal 0 in this manner, \[x^{2}+y^{2}=0\] then its the quadratic equation. just moved up 3 and to the left 2.

Answer 49

Here's an x-y table so you can plot the solutions: