Show that $P(x,y)=0$ is an ellipse if $b^2-4ac<0$.












2














I tried the following:



I write the polynomial $P(x, y) = ax^2+bxy+cy^2+dx+ey+h$ in the form $P(x, y) = Ax^2 + Bx + C$ where $A$, $B$, and $C$ are polynomial functions of $y$. This $P(x, y) = Q(x)$ has the discriminant $Delta_x(y) = (b^2 − 4ac)y^2 + (2bd − 4ae)y + (d^2 − 4ah)$. The discriminant of the $Delta_x(y) = (b^2 − 4ac)y^2 + (2bd − 4ae)y + (d^2 − 4ah)$ is $Delta_0 = 16(a^2e^2-aebd+acd^2+ah(b^2-4ac))$.



My questions:



i. The book says that if $b^2-4ac<0$, one of the following occurs: $K_1={{y | Delta_x(y) ge 0}} = emptyset$, $K_2={{y | Delta_x(y) ge 0}} = {{y_0}}$, or there exist real numbers $alpha$ and $beta$ such that $K_3={{y | Delta_x(y) ge 0}} = {{alpha le y le beta}}$. How to show that only one of this three cases happens?



ii. For cases $K_1$ and $K_2$ it is easy to know that $P(x,y)=0$ is empty or a single point, respectively. But how case $K_3$ implies $P(x,y)=0$ to be ellipse? (for example it can be a part of a hyperbola too [the normal one $pi/2$ rotated] since the domain is an interval and for any point on domain there are two points $y$ fitting in the equation - also IF for the $P(x, y)$ there are only possible shapes: circle/ellipse, parabola and hyperbola then the only choice remains is circle/ellipse).










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    2














    I tried the following:



    I write the polynomial $P(x, y) = ax^2+bxy+cy^2+dx+ey+h$ in the form $P(x, y) = Ax^2 + Bx + C$ where $A$, $B$, and $C$ are polynomial functions of $y$. This $P(x, y) = Q(x)$ has the discriminant $Delta_x(y) = (b^2 − 4ac)y^2 + (2bd − 4ae)y + (d^2 − 4ah)$. The discriminant of the $Delta_x(y) = (b^2 − 4ac)y^2 + (2bd − 4ae)y + (d^2 − 4ah)$ is $Delta_0 = 16(a^2e^2-aebd+acd^2+ah(b^2-4ac))$.



    My questions:



    i. The book says that if $b^2-4ac<0$, one of the following occurs: $K_1={{y | Delta_x(y) ge 0}} = emptyset$, $K_2={{y | Delta_x(y) ge 0}} = {{y_0}}$, or there exist real numbers $alpha$ and $beta$ such that $K_3={{y | Delta_x(y) ge 0}} = {{alpha le y le beta}}$. How to show that only one of this three cases happens?



    ii. For cases $K_1$ and $K_2$ it is easy to know that $P(x,y)=0$ is empty or a single point, respectively. But how case $K_3$ implies $P(x,y)=0$ to be ellipse? (for example it can be a part of a hyperbola too [the normal one $pi/2$ rotated] since the domain is an interval and for any point on domain there are two points $y$ fitting in the equation - also IF for the $P(x, y)$ there are only possible shapes: circle/ellipse, parabola and hyperbola then the only choice remains is circle/ellipse).










    share|cite|improve this question



























      2












      2








      2


      0





      I tried the following:



      I write the polynomial $P(x, y) = ax^2+bxy+cy^2+dx+ey+h$ in the form $P(x, y) = Ax^2 + Bx + C$ where $A$, $B$, and $C$ are polynomial functions of $y$. This $P(x, y) = Q(x)$ has the discriminant $Delta_x(y) = (b^2 − 4ac)y^2 + (2bd − 4ae)y + (d^2 − 4ah)$. The discriminant of the $Delta_x(y) = (b^2 − 4ac)y^2 + (2bd − 4ae)y + (d^2 − 4ah)$ is $Delta_0 = 16(a^2e^2-aebd+acd^2+ah(b^2-4ac))$.



      My questions:



      i. The book says that if $b^2-4ac<0$, one of the following occurs: $K_1={{y | Delta_x(y) ge 0}} = emptyset$, $K_2={{y | Delta_x(y) ge 0}} = {{y_0}}$, or there exist real numbers $alpha$ and $beta$ such that $K_3={{y | Delta_x(y) ge 0}} = {{alpha le y le beta}}$. How to show that only one of this three cases happens?



      ii. For cases $K_1$ and $K_2$ it is easy to know that $P(x,y)=0$ is empty or a single point, respectively. But how case $K_3$ implies $P(x,y)=0$ to be ellipse? (for example it can be a part of a hyperbola too [the normal one $pi/2$ rotated] since the domain is an interval and for any point on domain there are two points $y$ fitting in the equation - also IF for the $P(x, y)$ there are only possible shapes: circle/ellipse, parabola and hyperbola then the only choice remains is circle/ellipse).










      share|cite|improve this question















      I tried the following:



      I write the polynomial $P(x, y) = ax^2+bxy+cy^2+dx+ey+h$ in the form $P(x, y) = Ax^2 + Bx + C$ where $A$, $B$, and $C$ are polynomial functions of $y$. This $P(x, y) = Q(x)$ has the discriminant $Delta_x(y) = (b^2 − 4ac)y^2 + (2bd − 4ae)y + (d^2 − 4ah)$. The discriminant of the $Delta_x(y) = (b^2 − 4ac)y^2 + (2bd − 4ae)y + (d^2 − 4ah)$ is $Delta_0 = 16(a^2e^2-aebd+acd^2+ah(b^2-4ac))$.



      My questions:



      i. The book says that if $b^2-4ac<0$, one of the following occurs: $K_1={{y | Delta_x(y) ge 0}} = emptyset$, $K_2={{y | Delta_x(y) ge 0}} = {{y_0}}$, or there exist real numbers $alpha$ and $beta$ such that $K_3={{y | Delta_x(y) ge 0}} = {{alpha le y le beta}}$. How to show that only one of this three cases happens?



      ii. For cases $K_1$ and $K_2$ it is easy to know that $P(x,y)=0$ is empty or a single point, respectively. But how case $K_3$ implies $P(x,y)=0$ to be ellipse? (for example it can be a part of a hyperbola too [the normal one $pi/2$ rotated] since the domain is an interval and for any point on domain there are two points $y$ fitting in the equation - also IF for the $P(x, y)$ there are only possible shapes: circle/ellipse, parabola and hyperbola then the only choice remains is circle/ellipse).







      proof-verification algebraic-geometry conic-sections plane-curves






      share|cite|improve this question















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      share|cite|improve this question








      edited Nov 22 at 18:43

























      asked Nov 22 at 13:40









      72D

      543116




      543116






















          1 Answer
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          For i.



          Let $Y=Delta_x(y)$. Then, note that $Y=Delta_x(y)$ is the equation of a parabola. Moreover, the coefficient of $y^2$ is $b^2-4ac$ which is negative, so the parabola opens down. So, only one of the three cases happens.



          For ii.




          for example it can be a part of a hyperbola too [the normal one $pi/2$ rotated] since the domain is an interval and for any point on domain there are two points $y$ fitting in the equation




          It is impossible that $P(x,y)$ is a part of a hyperbola. Note that, in the first place, $P(x,y)=0$ has no restrictions on $x,y$. So, it is impossible that $P(x,y)=0$ is a part of something.
          In the case $K_3$ where there exist real numbers $alpha,beta$ such that $alphale yle beta$, the whole curve $P(x,y)=0$ is included in that interval. This immediately implies that $P(x,y)=0$ is neither a hyperbola nor a parabola.




          also IF for the $P(x, y)$ there are only possible shapes: circle/ellipse, parabola and hyperbola then the only choice remains is circle/ellipse).




          I think that in this context, it may be supposed that circles are a special case of ellipses.



          Anyway, in the case $K_3$, $P(x,y)=0$ can be a circle.



          If $a=1,b=0,c=1,d=-2,e=-2$ and $h=0$, then we have
          $$b^2-4ac=-4lt 0,qquad Delta_x(y) ge 0iff 1-sqrt 2le yle 1+sqrt 2$$
          and
          $$P(x,y)=0iff (x-1)^2+(y-1)^2=2$$
          which is the equation of a circle.





          Added :



          Suppose that $ax^2+bxy+cy^2$ becomes $AX^2+BXY+CY^2$ by the rotation of $theta$ :



          $$x=Xcostheta+Ysintheta,qquad y=-Xsintheta+Ycostheta$$



          Then, we get
          $$begin{align}A&=acos^2theta-bsinthetacostheta+csin^2theta
          \\B&=(a-c)sin(2theta)+bcos(2theta)
          \\C&=asin^2theta+bsinthetacostheta+csin^2thetaend{align}$$



          Since
          $$A+C=a+c,qquad A-C=(a-c)cos(2theta)-bsin(2theta)$$
          we get
          $$B^2-4AC=B^2-(A+C)^2+(A-C)^2=(a-c)^2+b^2-(a+c)^2=b^2-4ac$$



          So, when $B^2=0$, we get $$b^2-4ac=-4AC$$



          It follows that if $b^2-4aclt 0$, i.e. $ACgt 0$, i.e. either $Agt 0,Cgt 0$ or $Alt 0,Clt 0$, then the equation $$AX^2+0XY+CY^2+cdots =0$$ represents an ellipse.






          share|cite|improve this answer























          • ♥♥ LOVE YOU!! ♥♥
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          1 Answer
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          1 Answer
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          active

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          active

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          active

          oldest

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          2





          +50









          For i.



          Let $Y=Delta_x(y)$. Then, note that $Y=Delta_x(y)$ is the equation of a parabola. Moreover, the coefficient of $y^2$ is $b^2-4ac$ which is negative, so the parabola opens down. So, only one of the three cases happens.



          For ii.




          for example it can be a part of a hyperbola too [the normal one $pi/2$ rotated] since the domain is an interval and for any point on domain there are two points $y$ fitting in the equation




          It is impossible that $P(x,y)$ is a part of a hyperbola. Note that, in the first place, $P(x,y)=0$ has no restrictions on $x,y$. So, it is impossible that $P(x,y)=0$ is a part of something.
          In the case $K_3$ where there exist real numbers $alpha,beta$ such that $alphale yle beta$, the whole curve $P(x,y)=0$ is included in that interval. This immediately implies that $P(x,y)=0$ is neither a hyperbola nor a parabola.




          also IF for the $P(x, y)$ there are only possible shapes: circle/ellipse, parabola and hyperbola then the only choice remains is circle/ellipse).




          I think that in this context, it may be supposed that circles are a special case of ellipses.



          Anyway, in the case $K_3$, $P(x,y)=0$ can be a circle.



          If $a=1,b=0,c=1,d=-2,e=-2$ and $h=0$, then we have
          $$b^2-4ac=-4lt 0,qquad Delta_x(y) ge 0iff 1-sqrt 2le yle 1+sqrt 2$$
          and
          $$P(x,y)=0iff (x-1)^2+(y-1)^2=2$$
          which is the equation of a circle.





          Added :



          Suppose that $ax^2+bxy+cy^2$ becomes $AX^2+BXY+CY^2$ by the rotation of $theta$ :



          $$x=Xcostheta+Ysintheta,qquad y=-Xsintheta+Ycostheta$$



          Then, we get
          $$begin{align}A&=acos^2theta-bsinthetacostheta+csin^2theta
          \\B&=(a-c)sin(2theta)+bcos(2theta)
          \\C&=asin^2theta+bsinthetacostheta+csin^2thetaend{align}$$



          Since
          $$A+C=a+c,qquad A-C=(a-c)cos(2theta)-bsin(2theta)$$
          we get
          $$B^2-4AC=B^2-(A+C)^2+(A-C)^2=(a-c)^2+b^2-(a+c)^2=b^2-4ac$$



          So, when $B^2=0$, we get $$b^2-4ac=-4AC$$



          It follows that if $b^2-4aclt 0$, i.e. $ACgt 0$, i.e. either $Agt 0,Cgt 0$ or $Alt 0,Clt 0$, then the equation $$AX^2+0XY+CY^2+cdots =0$$ represents an ellipse.






          share|cite|improve this answer























          • ♥♥ LOVE YOU!! ♥♥
            – 72D
            Nov 27 at 0:25
















          2





          +50









          For i.



          Let $Y=Delta_x(y)$. Then, note that $Y=Delta_x(y)$ is the equation of a parabola. Moreover, the coefficient of $y^2$ is $b^2-4ac$ which is negative, so the parabola opens down. So, only one of the three cases happens.



          For ii.




          for example it can be a part of a hyperbola too [the normal one $pi/2$ rotated] since the domain is an interval and for any point on domain there are two points $y$ fitting in the equation




          It is impossible that $P(x,y)$ is a part of a hyperbola. Note that, in the first place, $P(x,y)=0$ has no restrictions on $x,y$. So, it is impossible that $P(x,y)=0$ is a part of something.
          In the case $K_3$ where there exist real numbers $alpha,beta$ such that $alphale yle beta$, the whole curve $P(x,y)=0$ is included in that interval. This immediately implies that $P(x,y)=0$ is neither a hyperbola nor a parabola.




          also IF for the $P(x, y)$ there are only possible shapes: circle/ellipse, parabola and hyperbola then the only choice remains is circle/ellipse).




          I think that in this context, it may be supposed that circles are a special case of ellipses.



          Anyway, in the case $K_3$, $P(x,y)=0$ can be a circle.



          If $a=1,b=0,c=1,d=-2,e=-2$ and $h=0$, then we have
          $$b^2-4ac=-4lt 0,qquad Delta_x(y) ge 0iff 1-sqrt 2le yle 1+sqrt 2$$
          and
          $$P(x,y)=0iff (x-1)^2+(y-1)^2=2$$
          which is the equation of a circle.





          Added :



          Suppose that $ax^2+bxy+cy^2$ becomes $AX^2+BXY+CY^2$ by the rotation of $theta$ :



          $$x=Xcostheta+Ysintheta,qquad y=-Xsintheta+Ycostheta$$



          Then, we get
          $$begin{align}A&=acos^2theta-bsinthetacostheta+csin^2theta
          \\B&=(a-c)sin(2theta)+bcos(2theta)
          \\C&=asin^2theta+bsinthetacostheta+csin^2thetaend{align}$$



          Since
          $$A+C=a+c,qquad A-C=(a-c)cos(2theta)-bsin(2theta)$$
          we get
          $$B^2-4AC=B^2-(A+C)^2+(A-C)^2=(a-c)^2+b^2-(a+c)^2=b^2-4ac$$



          So, when $B^2=0$, we get $$b^2-4ac=-4AC$$



          It follows that if $b^2-4aclt 0$, i.e. $ACgt 0$, i.e. either $Agt 0,Cgt 0$ or $Alt 0,Clt 0$, then the equation $$AX^2+0XY+CY^2+cdots =0$$ represents an ellipse.






          share|cite|improve this answer























          • ♥♥ LOVE YOU!! ♥♥
            – 72D
            Nov 27 at 0:25














          2





          +50







          2





          +50



          2




          +50




          For i.



          Let $Y=Delta_x(y)$. Then, note that $Y=Delta_x(y)$ is the equation of a parabola. Moreover, the coefficient of $y^2$ is $b^2-4ac$ which is negative, so the parabola opens down. So, only one of the three cases happens.



          For ii.




          for example it can be a part of a hyperbola too [the normal one $pi/2$ rotated] since the domain is an interval and for any point on domain there are two points $y$ fitting in the equation




          It is impossible that $P(x,y)$ is a part of a hyperbola. Note that, in the first place, $P(x,y)=0$ has no restrictions on $x,y$. So, it is impossible that $P(x,y)=0$ is a part of something.
          In the case $K_3$ where there exist real numbers $alpha,beta$ such that $alphale yle beta$, the whole curve $P(x,y)=0$ is included in that interval. This immediately implies that $P(x,y)=0$ is neither a hyperbola nor a parabola.




          also IF for the $P(x, y)$ there are only possible shapes: circle/ellipse, parabola and hyperbola then the only choice remains is circle/ellipse).




          I think that in this context, it may be supposed that circles are a special case of ellipses.



          Anyway, in the case $K_3$, $P(x,y)=0$ can be a circle.



          If $a=1,b=0,c=1,d=-2,e=-2$ and $h=0$, then we have
          $$b^2-4ac=-4lt 0,qquad Delta_x(y) ge 0iff 1-sqrt 2le yle 1+sqrt 2$$
          and
          $$P(x,y)=0iff (x-1)^2+(y-1)^2=2$$
          which is the equation of a circle.





          Added :



          Suppose that $ax^2+bxy+cy^2$ becomes $AX^2+BXY+CY^2$ by the rotation of $theta$ :



          $$x=Xcostheta+Ysintheta,qquad y=-Xsintheta+Ycostheta$$



          Then, we get
          $$begin{align}A&=acos^2theta-bsinthetacostheta+csin^2theta
          \\B&=(a-c)sin(2theta)+bcos(2theta)
          \\C&=asin^2theta+bsinthetacostheta+csin^2thetaend{align}$$



          Since
          $$A+C=a+c,qquad A-C=(a-c)cos(2theta)-bsin(2theta)$$
          we get
          $$B^2-4AC=B^2-(A+C)^2+(A-C)^2=(a-c)^2+b^2-(a+c)^2=b^2-4ac$$



          So, when $B^2=0$, we get $$b^2-4ac=-4AC$$



          It follows that if $b^2-4aclt 0$, i.e. $ACgt 0$, i.e. either $Agt 0,Cgt 0$ or $Alt 0,Clt 0$, then the equation $$AX^2+0XY+CY^2+cdots =0$$ represents an ellipse.






          share|cite|improve this answer














          For i.



          Let $Y=Delta_x(y)$. Then, note that $Y=Delta_x(y)$ is the equation of a parabola. Moreover, the coefficient of $y^2$ is $b^2-4ac$ which is negative, so the parabola opens down. So, only one of the three cases happens.



          For ii.




          for example it can be a part of a hyperbola too [the normal one $pi/2$ rotated] since the domain is an interval and for any point on domain there are two points $y$ fitting in the equation




          It is impossible that $P(x,y)$ is a part of a hyperbola. Note that, in the first place, $P(x,y)=0$ has no restrictions on $x,y$. So, it is impossible that $P(x,y)=0$ is a part of something.
          In the case $K_3$ where there exist real numbers $alpha,beta$ such that $alphale yle beta$, the whole curve $P(x,y)=0$ is included in that interval. This immediately implies that $P(x,y)=0$ is neither a hyperbola nor a parabola.




          also IF for the $P(x, y)$ there are only possible shapes: circle/ellipse, parabola and hyperbola then the only choice remains is circle/ellipse).




          I think that in this context, it may be supposed that circles are a special case of ellipses.



          Anyway, in the case $K_3$, $P(x,y)=0$ can be a circle.



          If $a=1,b=0,c=1,d=-2,e=-2$ and $h=0$, then we have
          $$b^2-4ac=-4lt 0,qquad Delta_x(y) ge 0iff 1-sqrt 2le yle 1+sqrt 2$$
          and
          $$P(x,y)=0iff (x-1)^2+(y-1)^2=2$$
          which is the equation of a circle.





          Added :



          Suppose that $ax^2+bxy+cy^2$ becomes $AX^2+BXY+CY^2$ by the rotation of $theta$ :



          $$x=Xcostheta+Ysintheta,qquad y=-Xsintheta+Ycostheta$$



          Then, we get
          $$begin{align}A&=acos^2theta-bsinthetacostheta+csin^2theta
          \\B&=(a-c)sin(2theta)+bcos(2theta)
          \\C&=asin^2theta+bsinthetacostheta+csin^2thetaend{align}$$



          Since
          $$A+C=a+c,qquad A-C=(a-c)cos(2theta)-bsin(2theta)$$
          we get
          $$B^2-4AC=B^2-(A+C)^2+(A-C)^2=(a-c)^2+b^2-(a+c)^2=b^2-4ac$$



          So, when $B^2=0$, we get $$b^2-4ac=-4AC$$



          It follows that if $b^2-4aclt 0$, i.e. $ACgt 0$, i.e. either $Agt 0,Cgt 0$ or $Alt 0,Clt 0$, then the equation $$AX^2+0XY+CY^2+cdots =0$$ represents an ellipse.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 25 at 6:12

























          answered Nov 25 at 5:05









          mathlove

          91.6k881214




          91.6k881214












          • ♥♥ LOVE YOU!! ♥♥
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            Nov 27 at 0:25


















          • ♥♥ LOVE YOU!! ♥♥
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          ♥♥ LOVE YOU!! ♥♥
          – 72D
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          ♥♥ LOVE YOU!! ♥♥
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          Nov 27 at 0:25


















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