Show that $P(x,y)=0$ is a parabola if $b^2-4ac=0$.
I tried the following:
I write the polynomial $P(x, y) = ax^2+bxy+cy^2+dx+ey+h$ in the form $P(x, y) = Ax2 + Bx + C$ where $A$, $B$, and $C$ are polynomial functions of $y$. This $P(x, y) = Q(x)$ has the discriminant $Delta_x(y) = (b^2 − 4ac)y^2 + (2bd − 4ae)y + (d^2 − 4ah)$. Letting $b^2-4ac=0$ makes $Delta_x(y)$ a linear function of $y$ and if $y ge frac{4ah -d^2}{2bd − 4ae}$ then for any of those $y$'s there is only one corresponding $x$. So IF for the $P(x, y)$ there are only four possible shapes: circle, ellipse, parabola and hyperbola then the only choice is parabola.
But the proof is not only complete but also doesn't look so rigorous. Are there any better proofs?
proof-verification algebraic-geometry conic-sections plane-curves
asked Nov 22 '18 at 12:26
72D
537116
add a comment |
I tried the following:
I write the polynomial $P(x, y) = ax^2+bxy+cy^2+dx+ey+h$ in the form $P(x, y) = Ax2 + Bx + C$ where $A$, $B$, and $C$ are polynomial functions of $y$. This $P(x, y) = Q(x)$ has the discriminant $Delta_x(y) = (b^2 − 4ac)y^2 + (2bd − 4ae)y + (d^2 − 4ah)$. Letting $b^2-4ac=0$ makes $Delta_x(y)$ a linear function of $y$ and if $y ge frac{4ah -d^2}{2bd − 4ae}$ then for any of those $y$'s there is only one corresponding $x$. So IF for the $P(x, y)$ there are only four possible shapes: circle, ellipse, parabola and hyperbola then the only choice is parabola.
But the proof is not only complete but also doesn't look so rigorous. Are there any better proofs?
proof-verification algebraic-geometry conic-sections plane-curves
asked Nov 22 '18 at 12:26
72D
537116
Strictly speaking, $P(x,y)=0$ is not necessarily a parabola when $b^2-4ac=0$. Take $a=b=c=d=0,e=h=1$. But as I've written in the answer to your another question, we can say that by a suitable rotation, $P(x,y)=0$ becomes an equation of the from $AX^2+CY^2+DX+EY+F=0$ with $AC=0$.
– mathlove
Nov 27 '18 at 5:26
@mathlove, evaluation of P(x,y)=0 by rotation method is much easier, but the book didn't introduced it (yet?). But after learning about it in your answer I could evaluated different cases too.
– 72D
Nov 27 '18 at 5:50
add a comment |
I tried the following:
I write the polynomial $P(x, y) = ax^2+bxy+cy^2+dx+ey+h$ in the form $P(x, y) = Ax2 + Bx + C$ where $A$, $B$, and $C$ are polynomial functions of $y$. This $P(x, y) = Q(x)$ has the discriminant $Delta_x(y) = (b^2 − 4ac)y^2 + (2bd − 4ae)y + (d^2 − 4ah)$. Letting $b^2-4ac=0$ makes $Delta_x(y)$ a linear function of $y$ and if $y ge frac{4ah -d^2}{2bd − 4ae}$ then for any of those $y$'s there is only one corresponding $x$. So IF for the $P(x, y)$ there are only four possible shapes: circle, ellipse, parabola and hyperbola then the only choice is parabola.
But the proof is not only complete but also doesn't look so rigorous. Are there any better proofs?
proof-verification algebraic-geometry conic-sections plane-curves
asked Nov 22 '18 at 12:26
72D
537116
I tried the following:
I write the polynomial $P(x, y) = ax^2+bxy+cy^2+dx+ey+h$ in the form $P(x, y) = Ax2 + Bx + C$ where $A$, $B$, and $C$ are polynomial functions of $y$. This $P(x, y) = Q(x)$ has the discriminant $Delta_x(y) = (b^2 − 4ac)y^2 + (2bd − 4ae)y + (d^2 − 4ah)$. Letting $b^2-4ac=0$ makes $Delta_x(y)$ a linear function of $y$ and if $y ge frac{4ah -d^2}{2bd − 4ae}$ then for any of those $y$'s there is only one corresponding $x$. So IF for the $P(x, y)$ there are only four possible shapes: circle, ellipse, parabola and hyperbola then the only choice is parabola.
But the proof is not only complete but also doesn't look so rigorous. Are there any better proofs?
proof-verification algebraic-geometry conic-sections plane-curves
proof-verification algebraic-geometry conic-sections plane-curves
asked Nov 22 '18 at 12:26
72D
537116
asked Nov 22 '18 at 12:26
72D
537116
edited Nov 22 '18 at 13:38
asked Nov 22 '18 at 12:26
72D
537116
asked Nov 22 '18 at 12:26
72D
537116
asked Nov 22 '18 at 12:26
72D
537116
537116
Strictly speaking, $P(x,y)=0$ is not necessarily a parabola when $b^2-4ac=0$. Take $a=b=c=d=0,e=h=1$. But as I've written in the answer to your another question, we can say that by a suitable rotation, $P(x,y)=0$ becomes an equation of the from $AX^2+CY^2+DX+EY+F=0$ with $AC=0$.
– mathlove
Nov 27 '18 at 5:26
@mathlove, evaluation of P(x,y)=0 by rotation method is much easier, but the book didn't introduced it (yet?). But after learning about it in your answer I could evaluated different cases too.
– 72D
Nov 27 '18 at 5:50
add a comment |
Strictly speaking, $P(x,y)=0$ is not necessarily a parabola when $b^2-4ac=0$. Take $a=b=c=d=0,e=h=1$. But as I've written in the answer to your another question, we can say that by a suitable rotation, $P(x,y)=0$ becomes an equation of the from $AX^2+CY^2+DX+EY+F=0$ with $AC=0$.
– mathlove
Nov 27 '18 at 5:26
@mathlove, evaluation of P(x,y)=0 by rotation method is much easier, but the book didn't introduced it (yet?). But after learning about it in your answer I could evaluated different cases too.
– 72D
Nov 27 '18 at 5:50
Strictly speaking, $P(x,y)=0$ is not necessarily a parabola when $b^2-4ac=0$. Take $a=b=c=d=0,e=h=1$. But as I've written in the answer to your another question, we can say that by a suitable rotation, $P(x,y)=0$ becomes an equation of the from $AX^2+CY^2+DX+EY+F=0$ with $AC=0$.
– mathlove
Nov 27 '18 at 5:26
Strictly speaking, $P(x,y)=0$ is not necessarily a parabola when $b^2-4ac=0$. Take $a=b=c=d=0,e=h=1$. But as I've written in the answer to your another question, we can say that by a suitable rotation, $P(x,y)=0$ becomes an equation of the from $AX^2+CY^2+DX+EY+F=0$ with $AC=0$.
– mathlove
Nov 27 '18 at 5:26
@mathlove, evaluation of P(x,y)=0 by rotation method is much easier, but the book didn't introduced it (yet?). But after learning about it in your answer I could evaluated different cases too.
– 72D
Nov 27 '18 at 5:50
@mathlove, evaluation of P(x,y)=0 by rotation method is much easier, but the book didn't introduced it (yet?). But after learning about it in your answer I could evaluated different cases too.
– 72D
Nov 27 '18 at 5:50
add a comment |
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Strictly speaking, $P(x,y)=0$ is not necessarily a parabola when $b^2-4ac=0$. Take $a=b=c=d=0,e=h=1$. But as I've written in the answer to your another question, we can say that by a suitable rotation, $P(x,y)=0$ becomes an equation of the from $AX^2+CY^2+DX+EY+F=0$ with $AC=0$.
– mathlove
Nov 27 '18 at 5:26
@mathlove, evaluation of P(x,y)=0 by rotation method is much easier, but the book didn't introduced it (yet?). But after learning about it in your answer I could evaluated different cases too.
– 72D
Nov 27 '18 at 5:50