Determine the equation of the second order curve
1
$begingroup$
I am trying to transform the equation of the second order curve to its canonical form and determine the type of the curve and plot its graph
$2x^2+y^2+4x-6y+11=0$
So I tried to use factoring polynomials
And got this form
$2(x+1)^2+(y-3)^2 = 0$ and actually I know $9 $ equations from the conic section but this doesn't look like one of them ! So I can't determine the type or graph it where is the mistake!
geometry conic-sections
asked Dec 24 '18 at 17:45
YolaYola
63
$endgroup$
add a comment |
1
$begingroup$
I am trying to transform the equation of the second order curve to its canonical form and determine the type of the curve and plot its graph
$2x^2+y^2+4x-6y+11=0$
So I tried to use factoring polynomials
And got this form
$2(x+1)^2+(y-3)^2 = 0$ and actually I know $9 $ equations from the conic section but this doesn't look like one of them ! So I can't determine the type or graph it where is the mistake!
geometry conic-sections
asked Dec 24 '18 at 17:45
YolaYola
63
$endgroup$
2
$begingroup$
The form you got has only $1 cdot y^2$ but the original has $2y^2.$
$endgroup$
– coffeemath
Dec 24 '18 at 17:53
1
$begingroup$
Is $-6x$ supposed to be $-6y?$ It appears so because you already have an $x$ term. Your standard form is not correct as the constant is $+11$ instead of $+12$. The form you show is just one point, as the sum of squares is only $0$ if they both are.
$endgroup$
– Ross Millikan
Dec 24 '18 at 18:06
$begingroup$
Yea it's-6y but and I solve it with -6y it's the same ...
$endgroup$
– Yola
Dec 24 '18 at 18:22
$begingroup$
It's -6y , And no the coefficient of y² is 1 not 2 if it's 2 it would be easier than to spit :)
$endgroup$
– Yola
Dec 24 '18 at 18:23
$begingroup$
You should edit the correction to the coefficient of $y^2$ into your post.
$endgroup$
– Ross Millikan
Dec 24 '18 at 18:38
add a comment |
1
1
1
$begingroup$
I am trying to transform the equation of the second order curve to its canonical form and determine the type of the curve and plot its graph
$2x^2+y^2+4x-6y+11=0$
So I tried to use factoring polynomials
And got this form
$2(x+1)^2+(y-3)^2 = 0$ and actually I know $9 $ equations from the conic section but this doesn't look like one of them ! So I can't determine the type or graph it where is the mistake!
geometry conic-sections
asked Dec 24 '18 at 17:45
YolaYola
63
$endgroup$
I am trying to transform the equation of the second order curve to its canonical form and determine the type of the curve and plot its graph
$2x^2+y^2+4x-6y+11=0$
So I tried to use factoring polynomials
And got this form
$2(x+1)^2+(y-3)^2 = 0$ and actually I know $9 $ equations from the conic section but this doesn't look like one of them ! So I can't determine the type or graph it where is the mistake!
geometry conic-sections
geometry conic-sections
asked Dec 24 '18 at 17:45
YolaYola
63
asked Dec 24 '18 at 17:45
YolaYola
63
edited Dec 24 '18 at 18:40
Yola
asked Dec 24 '18 at 17:45
YolaYola
63
asked Dec 24 '18 at 17:45
YolaYola
63
asked Dec 24 '18 at 17:45
YolaYola
63
63
2
$begingroup$
The form you got has only $1 cdot y^2$ but the original has $2y^2.$
$endgroup$
– coffeemath
Dec 24 '18 at 17:53
1
$begingroup$
Is $-6x$ supposed to be $-6y?$ It appears so because you already have an $x$ term. Your standard form is not correct as the constant is $+11$ instead of $+12$. The form you show is just one point, as the sum of squares is only $0$ if they both are.
$endgroup$
– Ross Millikan
Dec 24 '18 at 18:06
$begingroup$
Yea it's-6y but and I solve it with -6y it's the same ...
$endgroup$
– Yola
Dec 24 '18 at 18:22
$begingroup$
It's -6y , And no the coefficient of y² is 1 not 2 if it's 2 it would be easier than to spit :)
$endgroup$
– Yola
Dec 24 '18 at 18:23
$begingroup$
You should edit the correction to the coefficient of $y^2$ into your post.
$endgroup$
– Ross Millikan
Dec 24 '18 at 18:38
add a comment |
2
$begingroup$
The form you got has only $1 cdot y^2$ but the original has $2y^2.$
$endgroup$
– coffeemath
Dec 24 '18 at 17:53
1
$begingroup$
Is $-6x$ supposed to be $-6y?$ It appears so because you already have an $x$ term. Your standard form is not correct as the constant is $+11$ instead of $+12$. The form you show is just one point, as the sum of squares is only $0$ if they both are.
$endgroup$
– Ross Millikan
Dec 24 '18 at 18:06
$begingroup$
Yea it's-6y but and I solve it with -6y it's the same ...
$endgroup$
– Yola
Dec 24 '18 at 18:22
$begingroup$
It's -6y , And no the coefficient of y² is 1 not 2 if it's 2 it would be easier than to spit :)
$endgroup$
– Yola
Dec 24 '18 at 18:23
$begingroup$
You should edit the correction to the coefficient of $y^2$ into your post.
$endgroup$
– Ross Millikan
Dec 24 '18 at 18:38
2
2
$begingroup$
The form you got has only $1 cdot y^2$ but the original has $2y^2.$
$endgroup$
– coffeemath
Dec 24 '18 at 17:53
$begingroup$
The form you got has only $1 cdot y^2$ but the original has $2y^2.$
$endgroup$
– coffeemath
Dec 24 '18 at 17:53
1
1
$begingroup$
Is $-6x$ supposed to be $-6y?$ It appears so because you already have an $x$ term. Your standard form is not correct as the constant is $+11$ instead of $+12$. The form you show is just one point, as the sum of squares is only $0$ if they both are.
$endgroup$
– Ross Millikan
Dec 24 '18 at 18:06
$begingroup$
Is $-6x$ supposed to be $-6y?$ It appears so because you already have an $x$ term. Your standard form is not correct as the constant is $+11$ instead of $+12$. The form you show is just one point, as the sum of squares is only $0$ if they both are.
$endgroup$
– Ross Millikan
Dec 24 '18 at 18:06
$begingroup$
Yea it's-6y but and I solve it with -6y it's the same ...
$endgroup$
– Yola
Dec 24 '18 at 18:22
$begingroup$
Yea it's-6y but and I solve it with -6y it's the same ...
$endgroup$
– Yola
Dec 24 '18 at 18:22
$begingroup$
It's -6y , And no the coefficient of y² is 1 not 2 if it's 2 it would be easier than to spit :)
$endgroup$
– Yola
Dec 24 '18 at 18:23
$begingroup$
It's -6y , And no the coefficient of y² is 1 not 2 if it's 2 it would be easier than to spit :)
$endgroup$
– Yola
Dec 24 '18 at 18:23
$begingroup$
You should edit the correction to the coefficient of $y^2$ into your post.
$endgroup$
– Ross Millikan
Dec 24 '18 at 18:38
$begingroup$
You should edit the correction to the coefficient of $y^2$ into your post.
$endgroup$
– Ross Millikan
Dec 24 '18 at 18:38
add a comment |
1 Answer
1
active
oldest
votes
2
$begingroup$
Now that you have updated the constant term, you have the correct form $$2x^2+y^2+4x-6y+11=0\2(x+1)^2+(y-3)^2=0$$
When a sum of squares is zero, each square must be zero, so we can say
$$2(x+1)^2=0\(y-3)^2=0\x=-1\y=3$$
so the only solution is the point $(-1,3)$
answered Dec 24 '18 at 18:27
Ross MillikanRoss Millikan
295k23198371
$endgroup$
$begingroup$
Where that 1 came from
$endgroup$
– Yola
Dec 24 '18 at 18:35
$begingroup$
Which $2$? The one in front of $(x+1)^2$ is just like in your post. You can expand the squares in the last line to see it matches the first line.
$endgroup$
– Ross Millikan
Dec 24 '18 at 18:37
$begingroup$
We have 2(x²+2x+1-1)+ y² -6y+9-9+11=0. Then 2(x+1)²+(y-3)²-11+11= 0
$endgroup$
– Yola
Dec 24 '18 at 18:38
$begingroup$
But you have $12$ as the constant term, which I have copied. That is where the $+1$ comes from in the second line. It is what is left when the squared terms use up $11$
$endgroup$
– Ross Millikan
Dec 24 '18 at 18:39
1
$begingroup$
A single point is a degenerate form of a conic. You have that. The $+1$ and $-3$ just shift the center from the origin. You have $a^2=frac 12$. The right side being $0$ is what makes it a single point as I said.
$endgroup$
– Ross Millikan
Dec 24 '18 at 18:52
|
show 14 more comments
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1 Answer
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1 Answer
1
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2
$begingroup$
Now that you have updated the constant term, you have the correct form $$2x^2+y^2+4x-6y+11=0\2(x+1)^2+(y-3)^2=0$$
When a sum of squares is zero, each square must be zero, so we can say
$$2(x+1)^2=0\(y-3)^2=0\x=-1\y=3$$
so the only solution is the point $(-1,3)$
answered Dec 24 '18 at 18:27
Ross MillikanRoss Millikan
295k23198371
$endgroup$
$begingroup$
Where that 1 came from
$endgroup$
– Yola
Dec 24 '18 at 18:35
$begingroup$
Which $2$? The one in front of $(x+1)^2$ is just like in your post. You can expand the squares in the last line to see it matches the first line.
$endgroup$
– Ross Millikan
Dec 24 '18 at 18:37
$begingroup$
We have 2(x²+2x+1-1)+ y² -6y+9-9+11=0. Then 2(x+1)²+(y-3)²-11+11= 0
$endgroup$
– Yola
Dec 24 '18 at 18:38
$begingroup$
But you have $12$ as the constant term, which I have copied. That is where the $+1$ comes from in the second line. It is what is left when the squared terms use up $11$
$endgroup$
– Ross Millikan
Dec 24 '18 at 18:39
1
$begingroup$
A single point is a degenerate form of a conic. You have that. The $+1$ and $-3$ just shift the center from the origin. You have $a^2=frac 12$. The right side being $0$ is what makes it a single point as I said.
$endgroup$
– Ross Millikan
Dec 24 '18 at 18:52
|
show 14 more comments
2
$begingroup$
Now that you have updated the constant term, you have the correct form $$2x^2+y^2+4x-6y+11=0\2(x+1)^2+(y-3)^2=0$$
When a sum of squares is zero, each square must be zero, so we can say
$$2(x+1)^2=0\(y-3)^2=0\x=-1\y=3$$
so the only solution is the point $(-1,3)$
answered Dec 24 '18 at 18:27
Ross MillikanRoss Millikan
295k23198371
$endgroup$
$begingroup$
Where that 1 came from
$endgroup$
– Yola
Dec 24 '18 at 18:35
$begingroup$
Which $2$? The one in front of $(x+1)^2$ is just like in your post. You can expand the squares in the last line to see it matches the first line.
$endgroup$
– Ross Millikan
Dec 24 '18 at 18:37
$begingroup$
We have 2(x²+2x+1-1)+ y² -6y+9-9+11=0. Then 2(x+1)²+(y-3)²-11+11= 0
$endgroup$
– Yola
Dec 24 '18 at 18:38
$begingroup$
But you have $12$ as the constant term, which I have copied. That is where the $+1$ comes from in the second line. It is what is left when the squared terms use up $11$
$endgroup$
– Ross Millikan
Dec 24 '18 at 18:39
1
$begingroup$
A single point is a degenerate form of a conic. You have that. The $+1$ and $-3$ just shift the center from the origin. You have $a^2=frac 12$. The right side being $0$ is what makes it a single point as I said.
$endgroup$
– Ross Millikan
Dec 24 '18 at 18:52
|
show 14 more comments
2
2
2
$begingroup$
Now that you have updated the constant term, you have the correct form $$2x^2+y^2+4x-6y+11=0\2(x+1)^2+(y-3)^2=0$$
When a sum of squares is zero, each square must be zero, so we can say
$$2(x+1)^2=0\(y-3)^2=0\x=-1\y=3$$
so the only solution is the point $(-1,3)$
answered Dec 24 '18 at 18:27
Ross MillikanRoss Millikan
295k23198371
$endgroup$
Now that you have updated the constant term, you have the correct form $$2x^2+y^2+4x-6y+11=0\2(x+1)^2+(y-3)^2=0$$
When a sum of squares is zero, each square must be zero, so we can say
$$2(x+1)^2=0\(y-3)^2=0\x=-1\y=3$$
so the only solution is the point $(-1,3)$
answered Dec 24 '18 at 18:27
Ross MillikanRoss Millikan
295k23198371
edited Dec 24 '18 at 19:46
answered Dec 24 '18 at 18:27
Ross MillikanRoss Millikan
295k23198371
answered Dec 24 '18 at 18:27
Ross MillikanRoss Millikan
295k23198371
answered Dec 24 '18 at 18:27
Ross MillikanRoss Millikan
295k23198371
295k23198371
$begingroup$
Where that 1 came from
$endgroup$
– Yola
Dec 24 '18 at 18:35
$begingroup$
Which $2$? The one in front of $(x+1)^2$ is just like in your post. You can expand the squares in the last line to see it matches the first line.
$endgroup$
– Ross Millikan
Dec 24 '18 at 18:37
$begingroup$
We have 2(x²+2x+1-1)+ y² -6y+9-9+11=0. Then 2(x+1)²+(y-3)²-11+11= 0
$endgroup$
– Yola
Dec 24 '18 at 18:38
$begingroup$
But you have $12$ as the constant term, which I have copied. That is where the $+1$ comes from in the second line. It is what is left when the squared terms use up $11$
$endgroup$
– Ross Millikan
Dec 24 '18 at 18:39
1
$begingroup$
A single point is a degenerate form of a conic. You have that. The $+1$ and $-3$ just shift the center from the origin. You have $a^2=frac 12$. The right side being $0$ is what makes it a single point as I said.
$endgroup$
– Ross Millikan
Dec 24 '18 at 18:52
|
show 14 more comments
$begingroup$
Where that 1 came from
$endgroup$
– Yola
Dec 24 '18 at 18:35
$begingroup$
Which $2$? The one in front of $(x+1)^2$ is just like in your post. You can expand the squares in the last line to see it matches the first line.
$endgroup$
– Ross Millikan
Dec 24 '18 at 18:37
$begingroup$
We have 2(x²+2x+1-1)+ y² -6y+9-9+11=0. Then 2(x+1)²+(y-3)²-11+11= 0
$endgroup$
– Yola
Dec 24 '18 at 18:38
$begingroup$
But you have $12$ as the constant term, which I have copied. That is where the $+1$ comes from in the second line. It is what is left when the squared terms use up $11$
$endgroup$
– Ross Millikan
Dec 24 '18 at 18:39
1
$begingroup$
A single point is a degenerate form of a conic. You have that. The $+1$ and $-3$ just shift the center from the origin. You have $a^2=frac 12$. The right side being $0$ is what makes it a single point as I said.
$endgroup$
– Ross Millikan
Dec 24 '18 at 18:52
$begingroup$
Where that 1 came from
$endgroup$
– Yola
Dec 24 '18 at 18:35
$begingroup$
Where that 1 came from
$endgroup$
– Yola
Dec 24 '18 at 18:35
$begingroup$
Which $2$? The one in front of $(x+1)^2$ is just like in your post. You can expand the squares in the last line to see it matches the first line.
$endgroup$
– Ross Millikan
Dec 24 '18 at 18:37
$begingroup$
Which $2$? The one in front of $(x+1)^2$ is just like in your post. You can expand the squares in the last line to see it matches the first line.
$endgroup$
– Ross Millikan
Dec 24 '18 at 18:37
$begingroup$
We have 2(x²+2x+1-1)+ y² -6y+9-9+11=0. Then 2(x+1)²+(y-3)²-11+11= 0
$endgroup$
– Yola
Dec 24 '18 at 18:38
$begingroup$
We have 2(x²+2x+1-1)+ y² -6y+9-9+11=0. Then 2(x+1)²+(y-3)²-11+11= 0
$endgroup$
– Yola
Dec 24 '18 at 18:38
$begingroup$
But you have $12$ as the constant term, which I have copied. That is where the $+1$ comes from in the second line. It is what is left when the squared terms use up $11$
$endgroup$
– Ross Millikan
Dec 24 '18 at 18:39
$begingroup$
But you have $12$ as the constant term, which I have copied. That is where the $+1$ comes from in the second line. It is what is left when the squared terms use up $11$
$endgroup$
– Ross Millikan
Dec 24 '18 at 18:39
1
1
$begingroup$
A single point is a degenerate form of a conic. You have that. The $+1$ and $-3$ just shift the center from the origin. You have $a^2=frac 12$. The right side being $0$ is what makes it a single point as I said.
$endgroup$
– Ross Millikan
Dec 24 '18 at 18:52
$begingroup$
A single point is a degenerate form of a conic. You have that. The $+1$ and $-3$ just shift the center from the origin. You have $a^2=frac 12$. The right side being $0$ is what makes it a single point as I said.
$endgroup$
– Ross Millikan
Dec 24 '18 at 18:52
|
show 14 more comments
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2
$begingroup$
The form you got has only $1 cdot y^2$ but the original has $2y^2.$
$endgroup$
– coffeemath
Dec 24 '18 at 17:53
1
$begingroup$
Is $-6x$ supposed to be $-6y?$ It appears so because you already have an $x$ term. Your standard form is not correct as the constant is $+11$ instead of $+12$. The form you show is just one point, as the sum of squares is only $0$ if they both are.
$endgroup$
– Ross Millikan
Dec 24 '18 at 18:06
$begingroup$
Yea it's-6y but and I solve it with -6y it's the same ...
$endgroup$
– Yola
Dec 24 '18 at 18:22
$begingroup$
It's -6y , And no the coefficient of y² is 1 not 2 if it's 2 it would be easier than to spit :)
$endgroup$
– Yola
Dec 24 '18 at 18:23
$begingroup$
You should edit the correction to the coefficient of $y^2$ into your post.
$endgroup$
– Ross Millikan
Dec 24 '18 at 18:38