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1

$begingroup$

An illustration required me to find out the area of the curve $$5x^2 + 6xy + 2y^2 + 7x + +6y + 6 = 0 $$

They proceeded to solve the equation as a quadratic obtaining $y_1$ and $y_2$ as the two branches.

Why are $y_1$ and $y_2$ as depicted in the diagram?

^{if it isn't too broad what does it mean to solve a quadratic equation for any conic section (parabola, ellipse, hyperbola, circle.}

elliptic-curves

share|cite|improve this question

edited Mar 15 at 6:37

CaptainQuestion

asked Mar 15 at 6:28

CaptainQuestionCaptainQuestion

1769

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See this article in $colorredWikipedia$
  $endgroup$
  – steven gregory
  Mar 15 at 6:43
  
  

  
  
  
  

add a comment | 

1

$begingroup$

An illustration required me to find out the area of the curve $$5x^2 + 6xy + 2y^2 + 7x + +6y + 6 = 0 $$

They proceeded to solve the equation as a quadratic obtaining $y_1$ and $y_2$ as the two branches.

Why are $y_1$ and $y_2$ as depicted in the diagram?

^{if it isn't too broad what does it mean to solve a quadratic equation for any conic section (parabola, ellipse, hyperbola, circle.}

elliptic-curves

share|cite|improve this question

edited Mar 15 at 6:37

CaptainQuestion

asked Mar 15 at 6:28

CaptainQuestionCaptainQuestion

1769

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See this article in $colorredWikipedia$
  $endgroup$
  – steven gregory
  Mar 15 at 6:43
  
  

  
  
  
  

add a comment | 

$begingroup$

An illustration required me to find out the area of the curve $$5x^2 + 6xy + 2y^2 + 7x + +6y + 6 = 0 $$

They proceeded to solve the equation as a quadratic obtaining $y_1$ and $y_2$ as the two branches.

Why are $y_1$ and $y_2$ as depicted in the diagram?

^{if it isn't too broad what does it mean to solve a quadratic equation for any conic section (parabola, ellipse, hyperbola, circle.}

elliptic-curves

share|cite|improve this question

edited Mar 15 at 6:37

CaptainQuestion

asked Mar 15 at 6:28

CaptainQuestionCaptainQuestion

1769

$endgroup$

share|cite|improve this question

edited Mar 15 at 6:37

CaptainQuestion

asked Mar 15 at 6:28

CaptainQuestionCaptainQuestion

1769

share|cite|improve this question

edited Mar 15 at 6:37

CaptainQuestion

asked Mar 15 at 6:28

CaptainQuestionCaptainQuestion

1769

asked Mar 15 at 6:28

CaptainQuestionCaptainQuestion

1769

asked Mar 15 at 6:28

CaptainQuestionCaptainQuestion

1769

2 Answers
2

active

oldest

votes

2

$begingroup$

The point of this is to write $y$ as a function of $x$ - well, actually, two functions. So, then, we treat $x$ as a constant and sort everything by powers of $y$:
beginalign* 0 &= 5x^2+6xy+2y^2+7x+6y+6\
0 &= 2y^2 + (6xy+6y) + (5x^2+7x+6)\
0 &= 2y^2 + (6x+6)y + (5x^2+7x+6)endalign*
Since we're treating $x$ as a constant, that's just a quadratic equation in $y$ - and we know how to solve quadratic equations:
beginalign*y &= frac-(6x+6)pmsqrt(6x+6)^2-4cdot 2cdot (5x^2+7x+6)2cdot 2\
&= frac-6(x+1)pmsqrt36x^2+72x+36-40x^2-56x-484\
&= frac-6(x+1)pmsqrt-4x^2+16x-124\
y &= frac-3(x+1)pmsqrt(x-1)(3-x)2endalign*
There it is. That quadratic equation has a $pm$ sign in it, and both choices for the square root are viable. If we take the $+$ sign, we get a larger value of $y$, for the top curve $y_1$ drawn in pink. If we take the $-$ sign, we get a smaller value of $y$, for the bottom curve $y_2$ drawn in green.

Similar reasoning applies to any conic section; if we treat one variable as a constant, the equation becomes a linear or quadratic equation in the remaining variable, which we can solve explicitly.

share|cite|improve this answer

answered Mar 15 at 6:42

jmerryjmerry

15.6k1632

$endgroup$

add a comment | 

1

$begingroup$

I get
$$y_1=-fracsqrt-x^2+4 x-3+3 x+32$$
$$y_2=fracsqrt-x^2+4 x-3-3 x-32$$
Area of the ellipse is
$$S=int_1^3(y_2-y_1)dx\=int_1^3sqrt-x^2+4x-3dx=fracpi2$$

share|cite|improve this answer

edited Mar 15 at 7:31

answered Mar 15 at 7:22

Aleksas DomarkasAleksas Domarkas

1,54817

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This doesn’t really answer the question. The O.P. isn’t asking about the value of the area, but about the decomposition of the curve.
  $endgroup$
  – amd
  Mar 15 at 20:38
  

  
  
  
  

add a comment | 

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2

$begingroup$

The point of this is to write $y$ as a function of $x$ - well, actually, two functions. So, then, we treat $x$ as a constant and sort everything by powers of $y$:
beginalign* 0 &= 5x^2+6xy+2y^2+7x+6y+6\
0 &= 2y^2 + (6xy+6y) + (5x^2+7x+6)\
0 &= 2y^2 + (6x+6)y + (5x^2+7x+6)endalign*
Since we're treating $x$ as a constant, that's just a quadratic equation in $y$ - and we know how to solve quadratic equations:
beginalign*y &= frac-(6x+6)pmsqrt(6x+6)^2-4cdot 2cdot (5x^2+7x+6)2cdot 2\
&= frac-6(x+1)pmsqrt36x^2+72x+36-40x^2-56x-484\
&= frac-6(x+1)pmsqrt-4x^2+16x-124\
y &= frac-3(x+1)pmsqrt(x-1)(3-x)2endalign*
There it is. That quadratic equation has a $pm$ sign in it, and both choices for the square root are viable. If we take the $+$ sign, we get a larger value of $y$, for the top curve $y_1$ drawn in pink. If we take the $-$ sign, we get a smaller value of $y$, for the bottom curve $y_2$ drawn in green.

Similar reasoning applies to any conic section; if we treat one variable as a constant, the equation becomes a linear or quadratic equation in the remaining variable, which we can solve explicitly.

share|cite|improve this answer

answered Mar 15 at 6:42

jmerryjmerry

15.6k1632

$endgroup$

add a comment | 

2

$begingroup$

The point of this is to write $y$ as a function of $x$ - well, actually, two functions. So, then, we treat $x$ as a constant and sort everything by powers of $y$:
beginalign* 0 &= 5x^2+6xy+2y^2+7x+6y+6\
0 &= 2y^2 + (6xy+6y) + (5x^2+7x+6)\
0 &= 2y^2 + (6x+6)y + (5x^2+7x+6)endalign*
Since we're treating $x$ as a constant, that's just a quadratic equation in $y$ - and we know how to solve quadratic equations:
beginalign*y &= frac-(6x+6)pmsqrt(6x+6)^2-4cdot 2cdot (5x^2+7x+6)2cdot 2\
&= frac-6(x+1)pmsqrt36x^2+72x+36-40x^2-56x-484\
&= frac-6(x+1)pmsqrt-4x^2+16x-124\
y &= frac-3(x+1)pmsqrt(x-1)(3-x)2endalign*
There it is. That quadratic equation has a $pm$ sign in it, and both choices for the square root are viable. If we take the $+$ sign, we get a larger value of $y$, for the top curve $y_1$ drawn in pink. If we take the $-$ sign, we get a smaller value of $y$, for the bottom curve $y_2$ drawn in green.

Similar reasoning applies to any conic section; if we treat one variable as a constant, the equation becomes a linear or quadratic equation in the remaining variable, which we can solve explicitly.

share|cite|improve this answer

answered Mar 15 at 6:42

jmerryjmerry

15.6k1632

$endgroup$

add a comment | 

$begingroup$

The point of this is to write $y$ as a function of $x$ - well, actually, two functions. So, then, we treat $x$ as a constant and sort everything by powers of $y$:
beginalign* 0 &= 5x^2+6xy+2y^2+7x+6y+6\
0 &= 2y^2 + (6xy+6y) + (5x^2+7x+6)\
0 &= 2y^2 + (6x+6)y + (5x^2+7x+6)endalign*
Since we're treating $x$ as a constant, that's just a quadratic equation in $y$ - and we know how to solve quadratic equations:
beginalign*y &= frac-(6x+6)pmsqrt(6x+6)^2-4cdot 2cdot (5x^2+7x+6)2cdot 2\
&= frac-6(x+1)pmsqrt36x^2+72x+36-40x^2-56x-484\
&= frac-6(x+1)pmsqrt-4x^2+16x-124\
y &= frac-3(x+1)pmsqrt(x-1)(3-x)2endalign*
There it is. That quadratic equation has a $pm$ sign in it, and both choices for the square root are viable. If we take the $+$ sign, we get a larger value of $y$, for the top curve $y_1$ drawn in pink. If we take the $-$ sign, we get a smaller value of $y$, for the bottom curve $y_2$ drawn in green.

Similar reasoning applies to any conic section; if we treat one variable as a constant, the equation becomes a linear or quadratic equation in the remaining variable, which we can solve explicitly.

share|cite|improve this answer

answered Mar 15 at 6:42

jmerryjmerry

15.6k1632

$endgroup$

share|cite|improve this answer

answered Mar 15 at 6:42

jmerryjmerry

15.6k1632

answered Mar 15 at 6:42

jmerryjmerry

15.6k1632

answered Mar 15 at 6:42

jmerryjmerry

15.6k1632

1

$begingroup$

I get
$$y_1=-fracsqrt-x^2+4 x-3+3 x+32$$
$$y_2=fracsqrt-x^2+4 x-3-3 x-32$$
Area of the ellipse is
$$S=int_1^3(y_2-y_1)dx\=int_1^3sqrt-x^2+4x-3dx=fracpi2$$

share|cite|improve this answer

edited Mar 15 at 7:31

answered Mar 15 at 7:22

Aleksas DomarkasAleksas Domarkas

1,54817

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This doesn’t really answer the question. The O.P. isn’t asking about the value of the area, but about the decomposition of the curve.
  $endgroup$
  – amd
  Mar 15 at 20:38
  

  
  
  
  

add a comment | 

1

$begingroup$

I get
$$y_1=-fracsqrt-x^2+4 x-3+3 x+32$$
$$y_2=fracsqrt-x^2+4 x-3-3 x-32$$
Area of the ellipse is
$$S=int_1^3(y_2-y_1)dx\=int_1^3sqrt-x^2+4x-3dx=fracpi2$$

share|cite|improve this answer

edited Mar 15 at 7:31

answered Mar 15 at 7:22

Aleksas DomarkasAleksas Domarkas

1,54817

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This doesn’t really answer the question. The O.P. isn’t asking about the value of the area, but about the decomposition of the curve.
  $endgroup$
  – amd
  Mar 15 at 20:38
  

  
  
  
  

add a comment | 

$begingroup$

I get
$$y_1=-fracsqrt-x^2+4 x-3+3 x+32$$
$$y_2=fracsqrt-x^2+4 x-3-3 x-32$$
Area of the ellipse is
$$S=int_1^3(y_2-y_1)dx\=int_1^3sqrt-x^2+4x-3dx=fracpi2$$

share|cite|improve this answer

edited Mar 15 at 7:31

answered Mar 15 at 7:22

Aleksas DomarkasAleksas Domarkas

1,54817

$endgroup$

share|cite|improve this answer

edited Mar 15 at 7:31

answered Mar 15 at 7:22

Aleksas DomarkasAleksas Domarkas

1,54817

answered Mar 15 at 7:22

Aleksas DomarkasAleksas Domarkas

1,54817

answered Mar 15 at 7:22

Aleksas DomarkasAleksas Domarkas

1,54817