Finding the curve from the given equation. [closed]
The equation given is as follows :
$X^{4}-Y^{4}+2Y^{3}+2XY(Y-X)=0$.
We are asked to find out the type of curve represented by the above equation.I dont remember the options well.But it was like
$1$.equation of $2$ circles
$2$.equation of $2$ lines
$3$.equation of a circle and line
Can someone help me to solve this ?
geometry
asked Nov 23 at 5:08
ananda krishnan
33
closed as off-topic by Saad, Christopher, GNUSupporter 8964民主女神 地下教會, A. Pongrácz, Gibbs Nov 23 at 17:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, Christopher, A. Pongrácz, Gibbs
If this question can be reworded to fit the rules in the help center, please edit the question.
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The equation given is as follows :
$X^{4}-Y^{4}+2Y^{3}+2XY(Y-X)=0$.
We are asked to find out the type of curve represented by the above equation.I dont remember the options well.But it was like
$1$.equation of $2$ circles
$2$.equation of $2$ lines
$3$.equation of a circle and line
Can someone help me to solve this ?
geometry
asked Nov 23 at 5:08
ananda krishnan
33
closed as off-topic by Saad, Christopher, GNUSupporter 8964民主女神 地下教會, A. Pongrácz, Gibbs Nov 23 at 17:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, Christopher, A. Pongrácz, Gibbs
If this question can be reworded to fit the rules in the help center, please edit the question.
Are you sure you wrote that down properly? For whatever it's worth, I just graphed that curve in Mathematica, and it doesn't appear to be any of those things.
– John Barber
Nov 23 at 5:30
What are you getting?I am not very sure about the options.Forget the options.Please tell me what you are getting.
– ananda krishnan
Nov 23 at 5:40
1
I'm getting two weird almost-lines that become curvy shapes near the origin. The reason I asked if you wrote it down correctly is that if you had an extra factor of $-2X^3$ in there, you'd get $X^4 - Y^4 + 2(Y^3 - X^3) + 2 X Y (Y - X) = (X - Y) (X + Y) (-2 X + X^2 - 2 Y + Y^2) = 0$, which is exactly the equation for two lines and one circle.
– John Barber
Nov 23 at 5:49
Anyway thanx for the reply.I am happy with your reply.Maybe the question was wrong.
– ananda krishnan
Nov 23 at 5:58
Can yu write the full steps on how you coverted that equation into this form of 2 lines and a circle?
– ananda krishnan
Nov 23 at 6:31
|
show 1 more comment
The equation given is as follows :
$X^{4}-Y^{4}+2Y^{3}+2XY(Y-X)=0$.
We are asked to find out the type of curve represented by the above equation.I dont remember the options well.But it was like
$1$.equation of $2$ circles
$2$.equation of $2$ lines
$3$.equation of a circle and line
Can someone help me to solve this ?
geometry
asked Nov 23 at 5:08
ananda krishnan
33
The equation given is as follows :
$X^{4}-Y^{4}+2Y^{3}+2XY(Y-X)=0$.
We are asked to find out the type of curve represented by the above equation.I dont remember the options well.But it was like
$1$.equation of $2$ circles
$2$.equation of $2$ lines
$3$.equation of a circle and line
Can someone help me to solve this ?
geometry
geometry
asked Nov 23 at 5:08
ananda krishnan
33
asked Nov 23 at 5:08
ananda krishnan
33
edited Nov 23 at 5:14
asked Nov 23 at 5:08
ananda krishnan
33
asked Nov 23 at 5:08
ananda krishnan
33
asked Nov 23 at 5:08
ananda krishnan
33
33
closed as off-topic by Saad, Christopher, GNUSupporter 8964民主女神 地下教會, A. Pongrácz, Gibbs Nov 23 at 17:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, Christopher, A. Pongrácz, Gibbs
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, Christopher, GNUSupporter 8964民主女神 地下教會, A. Pongrácz, Gibbs Nov 23 at 17:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, Christopher, A. Pongrácz, Gibbs
If this question can be reworded to fit the rules in the help center, please edit the question.
Are you sure you wrote that down properly? For whatever it's worth, I just graphed that curve in Mathematica, and it doesn't appear to be any of those things.
– John Barber
Nov 23 at 5:30
What are you getting?I am not very sure about the options.Forget the options.Please tell me what you are getting.
– ananda krishnan
Nov 23 at 5:40
1
I'm getting two weird almost-lines that become curvy shapes near the origin. The reason I asked if you wrote it down correctly is that if you had an extra factor of $-2X^3$ in there, you'd get $X^4 - Y^4 + 2(Y^3 - X^3) + 2 X Y (Y - X) = (X - Y) (X + Y) (-2 X + X^2 - 2 Y + Y^2) = 0$, which is exactly the equation for two lines and one circle.
– John Barber
Nov 23 at 5:49
Anyway thanx for the reply.I am happy with your reply.Maybe the question was wrong.
– ananda krishnan
Nov 23 at 5:58
Can yu write the full steps on how you coverted that equation into this form of 2 lines and a circle?
– ananda krishnan
Nov 23 at 6:31
|
show 1 more comment
Are you sure you wrote that down properly? For whatever it's worth, I just graphed that curve in Mathematica, and it doesn't appear to be any of those things.
– John Barber
Nov 23 at 5:30
What are you getting?I am not very sure about the options.Forget the options.Please tell me what you are getting.
– ananda krishnan
Nov 23 at 5:40
1
I'm getting two weird almost-lines that become curvy shapes near the origin. The reason I asked if you wrote it down correctly is that if you had an extra factor of $-2X^3$ in there, you'd get $X^4 - Y^4 + 2(Y^3 - X^3) + 2 X Y (Y - X) = (X - Y) (X + Y) (-2 X + X^2 - 2 Y + Y^2) = 0$, which is exactly the equation for two lines and one circle.
– John Barber
Nov 23 at 5:49
Anyway thanx for the reply.I am happy with your reply.Maybe the question was wrong.
– ananda krishnan
Nov 23 at 5:58
Can yu write the full steps on how you coverted that equation into this form of 2 lines and a circle?
– ananda krishnan
Nov 23 at 6:31
Are you sure you wrote that down properly? For whatever it's worth, I just graphed that curve in Mathematica, and it doesn't appear to be any of those things.
– John Barber
Nov 23 at 5:30
Are you sure you wrote that down properly? For whatever it's worth, I just graphed that curve in Mathematica, and it doesn't appear to be any of those things.
– John Barber
Nov 23 at 5:30
What are you getting?I am not very sure about the options.Forget the options.Please tell me what you are getting.
– ananda krishnan
Nov 23 at 5:40
What are you getting?I am not very sure about the options.Forget the options.Please tell me what you are getting.
– ananda krishnan
Nov 23 at 5:40
1
1
I'm getting two weird almost-lines that become curvy shapes near the origin. The reason I asked if you wrote it down correctly is that if you had an extra factor of $-2X^3$ in there, you'd get $X^4 - Y^4 + 2(Y^3 - X^3) + 2 X Y (Y - X) = (X - Y) (X + Y) (-2 X + X^2 - 2 Y + Y^2) = 0$, which is exactly the equation for two lines and one circle.
– John Barber
Nov 23 at 5:49
I'm getting two weird almost-lines that become curvy shapes near the origin. The reason I asked if you wrote it down correctly is that if you had an extra factor of $-2X^3$ in there, you'd get $X^4 - Y^4 + 2(Y^3 - X^3) + 2 X Y (Y - X) = (X - Y) (X + Y) (-2 X + X^2 - 2 Y + Y^2) = 0$, which is exactly the equation for two lines and one circle.
– John Barber
Nov 23 at 5:49
Anyway thanx for the reply.I am happy with your reply.Maybe the question was wrong.
– ananda krishnan
Nov 23 at 5:58
Anyway thanx for the reply.I am happy with your reply.Maybe the question was wrong.
– ananda krishnan
Nov 23 at 5:58
Can yu write the full steps on how you coverted that equation into this form of 2 lines and a circle?
– ananda krishnan
Nov 23 at 6:31
Can yu write the full steps on how you coverted that equation into this form of 2 lines and a circle?
– ananda krishnan
Nov 23 at 6:31
|
show 1 more comment
1 Answer
1
active
oldest
votes
You are indeed missing a factor of $2X^3$.
As for factoring here it is-
$$X^4-Y^4+2Y^3-2X^3+2XY(Y-X)=0$$
$$=(X^2+Y^2)(X^2-Y^2)+2(Y-X)(X^2+Y^2+XY)+2XY(Y-X)$$
$$=(X-Y)(X+Y)(X^2+Y^2)+2(Y-X)(X^2+Y^2+XY)+2XY(Y-X)$$
Factoring out $(Y-X)$
$$=(Y-X) [ (X+Y)(-X^2-Y^2)+2(X^2+Y^2+XY)+2XY ]$$
$$=(Y-X) [ (X+Y)(-X^2-Y^2)+2(X^2+Y^2+XY+XY) ]$$
$$=(Y-X) [ (X+Y)(-X^2-Y^2)+2(X^2+Y^2+2XY) ]$$
$$=(Y-X) [ (X+Y)(-X^2-Y^2)+2(X+Y)(X+Y) ]$$
Factoring out $(Y+X)$
$$=(Y-X)(Y+X)(-X^2-Y^2+X+Y)$$
Now either $x+y=0$ => $x=-y$
or $x-y=0$ => $x=y$ (equations of line)
or $(-X^2-Y^2+X+Y)=0$ (equation of circle)
answered Nov 23 at 15:01
Etotheipi
344
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You are indeed missing a factor of $2X^3$.
As for factoring here it is-
$$X^4-Y^4+2Y^3-2X^3+2XY(Y-X)=0$$
$$=(X^2+Y^2)(X^2-Y^2)+2(Y-X)(X^2+Y^2+XY)+2XY(Y-X)$$
$$=(X-Y)(X+Y)(X^2+Y^2)+2(Y-X)(X^2+Y^2+XY)+2XY(Y-X)$$
Factoring out $(Y-X)$
$$=(Y-X) [ (X+Y)(-X^2-Y^2)+2(X^2+Y^2+XY)+2XY ]$$
$$=(Y-X) [ (X+Y)(-X^2-Y^2)+2(X^2+Y^2+XY+XY) ]$$
$$=(Y-X) [ (X+Y)(-X^2-Y^2)+2(X^2+Y^2+2XY) ]$$
$$=(Y-X) [ (X+Y)(-X^2-Y^2)+2(X+Y)(X+Y) ]$$
Factoring out $(Y+X)$
$$=(Y-X)(Y+X)(-X^2-Y^2+X+Y)$$
Now either $x+y=0$ => $x=-y$
or $x-y=0$ => $x=y$ (equations of line)
or $(-X^2-Y^2+X+Y)=0$ (equation of circle)
answered Nov 23 at 15:01
Etotheipi
344
add a comment |
You are indeed missing a factor of $2X^3$.
As for factoring here it is-
$$X^4-Y^4+2Y^3-2X^3+2XY(Y-X)=0$$
$$=(X^2+Y^2)(X^2-Y^2)+2(Y-X)(X^2+Y^2+XY)+2XY(Y-X)$$
$$=(X-Y)(X+Y)(X^2+Y^2)+2(Y-X)(X^2+Y^2+XY)+2XY(Y-X)$$
Factoring out $(Y-X)$
$$=(Y-X) [ (X+Y)(-X^2-Y^2)+2(X^2+Y^2+XY)+2XY ]$$
$$=(Y-X) [ (X+Y)(-X^2-Y^2)+2(X^2+Y^2+XY+XY) ]$$
$$=(Y-X) [ (X+Y)(-X^2-Y^2)+2(X^2+Y^2+2XY) ]$$
$$=(Y-X) [ (X+Y)(-X^2-Y^2)+2(X+Y)(X+Y) ]$$
Factoring out $(Y+X)$
$$=(Y-X)(Y+X)(-X^2-Y^2+X+Y)$$
Now either $x+y=0$ => $x=-y$
or $x-y=0$ => $x=y$ (equations of line)
or $(-X^2-Y^2+X+Y)=0$ (equation of circle)
answered Nov 23 at 15:01
Etotheipi
344
add a comment |
You are indeed missing a factor of $2X^3$.
As for factoring here it is-
$$X^4-Y^4+2Y^3-2X^3+2XY(Y-X)=0$$
$$=(X^2+Y^2)(X^2-Y^2)+2(Y-X)(X^2+Y^2+XY)+2XY(Y-X)$$
$$=(X-Y)(X+Y)(X^2+Y^2)+2(Y-X)(X^2+Y^2+XY)+2XY(Y-X)$$
Factoring out $(Y-X)$
$$=(Y-X) [ (X+Y)(-X^2-Y^2)+2(X^2+Y^2+XY)+2XY ]$$
$$=(Y-X) [ (X+Y)(-X^2-Y^2)+2(X^2+Y^2+XY+XY) ]$$
$$=(Y-X) [ (X+Y)(-X^2-Y^2)+2(X^2+Y^2+2XY) ]$$
$$=(Y-X) [ (X+Y)(-X^2-Y^2)+2(X+Y)(X+Y) ]$$
Factoring out $(Y+X)$
$$=(Y-X)(Y+X)(-X^2-Y^2+X+Y)$$
Now either $x+y=0$ => $x=-y$
or $x-y=0$ => $x=y$ (equations of line)
or $(-X^2-Y^2+X+Y)=0$ (equation of circle)
answered Nov 23 at 15:01
Etotheipi
344
You are indeed missing a factor of $2X^3$.
As for factoring here it is-
$$X^4-Y^4+2Y^3-2X^3+2XY(Y-X)=0$$
$$=(X^2+Y^2)(X^2-Y^2)+2(Y-X)(X^2+Y^2+XY)+2XY(Y-X)$$
$$=(X-Y)(X+Y)(X^2+Y^2)+2(Y-X)(X^2+Y^2+XY)+2XY(Y-X)$$
Factoring out $(Y-X)$
$$=(Y-X) [ (X+Y)(-X^2-Y^2)+2(X^2+Y^2+XY)+2XY ]$$
$$=(Y-X) [ (X+Y)(-X^2-Y^2)+2(X^2+Y^2+XY+XY) ]$$
$$=(Y-X) [ (X+Y)(-X^2-Y^2)+2(X^2+Y^2+2XY) ]$$
$$=(Y-X) [ (X+Y)(-X^2-Y^2)+2(X+Y)(X+Y) ]$$
Factoring out $(Y+X)$
$$=(Y-X)(Y+X)(-X^2-Y^2+X+Y)$$
Now either $x+y=0$ => $x=-y$
or $x-y=0$ => $x=y$ (equations of line)
or $(-X^2-Y^2+X+Y)=0$ (equation of circle)
answered Nov 23 at 15:01
Etotheipi
344
answered Nov 23 at 15:01
Etotheipi
344
answered Nov 23 at 15:01
Etotheipi
344
answered Nov 23 at 15:01
Etotheipi
344
344
add a comment |
add a comment |
Are you sure you wrote that down properly? For whatever it's worth, I just graphed that curve in Mathematica, and it doesn't appear to be any of those things.
– John Barber
Nov 23 at 5:30
What are you getting?I am not very sure about the options.Forget the options.Please tell me what you are getting.
– ananda krishnan
Nov 23 at 5:40
1
I'm getting two weird almost-lines that become curvy shapes near the origin. The reason I asked if you wrote it down correctly is that if you had an extra factor of $-2X^3$ in there, you'd get $X^4 - Y^4 + 2(Y^3 - X^3) + 2 X Y (Y - X) = (X - Y) (X + Y) (-2 X + X^2 - 2 Y + Y^2) = 0$, which is exactly the equation for two lines and one circle.
– John Barber
Nov 23 at 5:49
Anyway thanx for the reply.I am happy with your reply.Maybe the question was wrong.
– ananda krishnan
Nov 23 at 5:58
Can yu write the full steps on how you coverted that equation into this form of 2 lines and a circle?
– ananda krishnan
Nov 23 at 6:31