Prove that the parabloas are mutually perpendicular.












0












$begingroup$


Given that two parabolas have the same focus with their axes of symmetry in opposite directions. Then I have to prove that the two intersect at right angles. As I think, because the axes are oppositely directed, the directrices must be rotated by an angle $π$ . So let their equations be $x+y+c=0$ and $x-y+c=0$. Let the common focus of the two parabolas be, for simplicity, the origin $(0,0)$. Then the two parabolas are:
$$F(x,y):=x^2+y^2-(x+y+c)^2=0$$
$$G(x,y):=x^2+y^2-(x-y+c)^2=0$$
Hence , the angle between the two parabolas comes out to be:
$$atan frac {F_xG_y - F_yG_x}{F_xG_x + F_yG_y}$$
$$= atan frac {2c(c+x)}{x^2+2cx+y^2}$$
Where am I getting wrong?
The subscripts stand for partial differentials.






analytic-geometry conic-sections







share|cite|improve this question









$endgroup$












  • $begingroup$
    What do you mean by "axes ... in opposite direction"?
    $endgroup$
    – user10354138
    Dec 9 '18 at 9:41










  • $begingroup$
    @user10354138, this means that the axes of symmetry of the two parabolas are antiparallel.
    $endgroup$
    – Awe Kumar Jha
    Dec 9 '18 at 9:47










  • $begingroup$
    By antiparallel is meant perpendicular?
    $endgroup$
    – coffeemath
    Dec 9 '18 at 9:52










  • $begingroup$
    @coffeemath, of course not. Antiparallel is parallel but rotated by 180 deg.
    $endgroup$
    – Awe Kumar Jha
    Dec 9 '18 at 10:01






  • 1




    $begingroup$
    But $xpm y-c=0$ are perpendicular, not (anti-)parallel.
    $endgroup$
    – user10354138
    Dec 9 '18 at 10:01


















0












$begingroup$


Given that two parabolas have the same focus with their axes of symmetry in opposite directions. Then I have to prove that the two intersect at right angles. As I think, because the axes are oppositely directed, the directrices must be rotated by an angle $π$ . So let their equations be $x+y+c=0$ and $x-y+c=0$. Let the common focus of the two parabolas be, for simplicity, the origin $(0,0)$. Then the two parabolas are:
$$F(x,y):=x^2+y^2-(x+y+c)^2=0$$
$$G(x,y):=x^2+y^2-(x-y+c)^2=0$$
Hence , the angle between the two parabolas comes out to be:
$$atan frac {F_xG_y - F_yG_x}{F_xG_x + F_yG_y}$$
$$= atan frac {2c(c+x)}{x^2+2cx+y^2}$$
Where am I getting wrong?
The subscripts stand for partial differentials.






analytic-geometry conic-sections







share|cite|improve this question









$endgroup$












  • $begingroup$
    What do you mean by "axes ... in opposite direction"?
    $endgroup$
    – user10354138
    Dec 9 '18 at 9:41










  • $begingroup$
    @user10354138, this means that the axes of symmetry of the two parabolas are antiparallel.
    $endgroup$
    – Awe Kumar Jha
    Dec 9 '18 at 9:47










  • $begingroup$
    By antiparallel is meant perpendicular?
    $endgroup$
    – coffeemath
    Dec 9 '18 at 9:52










  • $begingroup$
    @coffeemath, of course not. Antiparallel is parallel but rotated by 180 deg.
    $endgroup$
    – Awe Kumar Jha
    Dec 9 '18 at 10:01






  • 1




    $begingroup$
    But $xpm y-c=0$ are perpendicular, not (anti-)parallel.
    $endgroup$
    – user10354138
    Dec 9 '18 at 10:01
















0












0








0


0



$begingroup$


Given that two parabolas have the same focus with their axes of symmetry in opposite directions. Then I have to prove that the two intersect at right angles. As I think, because the axes are oppositely directed, the directrices must be rotated by an angle $π$ . So let their equations be $x+y+c=0$ and $x-y+c=0$. Let the common focus of the two parabolas be, for simplicity, the origin $(0,0)$. Then the two parabolas are:
$$F(x,y):=x^2+y^2-(x+y+c)^2=0$$
$$G(x,y):=x^2+y^2-(x-y+c)^2=0$$
Hence , the angle between the two parabolas comes out to be:
$$atan frac {F_xG_y - F_yG_x}{F_xG_x + F_yG_y}$$
$$= atan frac {2c(c+x)}{x^2+2cx+y^2}$$
Where am I getting wrong?
The subscripts stand for partial differentials.






analytic-geometry conic-sections







share|cite|improve this question









$endgroup$




Given that two parabolas have the same focus with their axes of symmetry in opposite directions. Then I have to prove that the two intersect at right angles. As I think, because the axes are oppositely directed, the directrices must be rotated by an angle $π$ . So let their equations be $x+y+c=0$ and $x-y+c=0$. Let the common focus of the two parabolas be, for simplicity, the origin $(0,0)$. Then the two parabolas are:
$$F(x,y):=x^2+y^2-(x+y+c)^2=0$$
$$G(x,y):=x^2+y^2-(x-y+c)^2=0$$
Hence , the angle between the two parabolas comes out to be:
$$atan frac {F_xG_y - F_yG_x}{F_xG_x + F_yG_y}$$
$$= atan frac {2c(c+x)}{x^2+2cx+y^2}$$
Where am I getting wrong?
The subscripts stand for partial differentials.






analytic-geometry conic-sections




analytic-geometry conic-sections






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 9 '18 at 9:13









Awe Kumar JhaAwe Kumar Jha

523113




523113












  • $begingroup$
    What do you mean by "axes ... in opposite direction"?
    $endgroup$
    – user10354138
    Dec 9 '18 at 9:41










  • $begingroup$
    @user10354138, this means that the axes of symmetry of the two parabolas are antiparallel.
    $endgroup$
    – Awe Kumar Jha
    Dec 9 '18 at 9:47










  • $begingroup$
    By antiparallel is meant perpendicular?
    $endgroup$
    – coffeemath
    Dec 9 '18 at 9:52










  • $begingroup$
    @coffeemath, of course not. Antiparallel is parallel but rotated by 180 deg.
    $endgroup$
    – Awe Kumar Jha
    Dec 9 '18 at 10:01






  • 1




    $begingroup$
    But $xpm y-c=0$ are perpendicular, not (anti-)parallel.
    $endgroup$
    – user10354138
    Dec 9 '18 at 10:01




















  • $begingroup$
    What do you mean by "axes ... in opposite direction"?
    $endgroup$
    – user10354138
    Dec 9 '18 at 9:41










  • $begingroup$
    @user10354138, this means that the axes of symmetry of the two parabolas are antiparallel.
    $endgroup$
    – Awe Kumar Jha
    Dec 9 '18 at 9:47










  • $begingroup$
    By antiparallel is meant perpendicular?
    $endgroup$
    – coffeemath
    Dec 9 '18 at 9:52










  • $begingroup$
    @coffeemath, of course not. Antiparallel is parallel but rotated by 180 deg.
    $endgroup$
    – Awe Kumar Jha
    Dec 9 '18 at 10:01






  • 1




    $begingroup$
    But $xpm y-c=0$ are perpendicular, not (anti-)parallel.
    $endgroup$
    – user10354138
    Dec 9 '18 at 10:01


















$begingroup$
What do you mean by "axes ... in opposite direction"?
$endgroup$
– user10354138
Dec 9 '18 at 9:41




$begingroup$
What do you mean by "axes ... in opposite direction"?
$endgroup$
– user10354138
Dec 9 '18 at 9:41












$begingroup$
@user10354138, this means that the axes of symmetry of the two parabolas are antiparallel.
$endgroup$
– Awe Kumar Jha
Dec 9 '18 at 9:47




$begingroup$
@user10354138, this means that the axes of symmetry of the two parabolas are antiparallel.
$endgroup$
– Awe Kumar Jha
Dec 9 '18 at 9:47












$begingroup$
By antiparallel is meant perpendicular?
$endgroup$
– coffeemath
Dec 9 '18 at 9:52




$begingroup$
By antiparallel is meant perpendicular?
$endgroup$
– coffeemath
Dec 9 '18 at 9:52












$begingroup$
@coffeemath, of course not. Antiparallel is parallel but rotated by 180 deg.
$endgroup$
– Awe Kumar Jha
Dec 9 '18 at 10:01




$begingroup$
@coffeemath, of course not. Antiparallel is parallel but rotated by 180 deg.
$endgroup$
– Awe Kumar Jha
Dec 9 '18 at 10:01




1




1




$begingroup$
But $xpm y-c=0$ are perpendicular, not (anti-)parallel.
$endgroup$
– user10354138
Dec 9 '18 at 10:01






$begingroup$
But $xpm y-c=0$ are perpendicular, not (anti-)parallel.
$endgroup$
– user10354138
Dec 9 '18 at 10:01












1 Answer
1






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oldest

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1












$begingroup$

If the parabolae are conguent: Suppose the directrices are $y=c$ and $y=-c$. Then the equations of the parabolae are:
$$Gamma_pm:=(x^2+y^2=(ymp c)^2)=(x^2pm2yc-c^2=0)$$
They intersect at $(pm c,0)$ and it is easy to see their tangents have slope $pm 1$ there.



If the parabolae are not congruent: the directrices $y=c_i$, $i=1,2$ gives
$$
Gamma_i:=(x^2+2yc_i-c_i^2=0)
$$
so $y=frac12(c_1+c_2)$ and $x=pmsqrt{-c_1c_2}$ (in particular, $c_1c_2<0$). Differentiating gives the slopes are $-frac{c_i}{x}$ and so multiplies to $-1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Oops, yes that is indeed not given. Corrected.
    $endgroup$
    – user10354138
    Dec 9 '18 at 15:47













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1 Answer
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1












$begingroup$

If the parabolae are conguent: Suppose the directrices are $y=c$ and $y=-c$. Then the equations of the parabolae are:
$$Gamma_pm:=(x^2+y^2=(ymp c)^2)=(x^2pm2yc-c^2=0)$$
They intersect at $(pm c,0)$ and it is easy to see their tangents have slope $pm 1$ there.



If the parabolae are not congruent: the directrices $y=c_i$, $i=1,2$ gives
$$
Gamma_i:=(x^2+2yc_i-c_i^2=0)
$$
so $y=frac12(c_1+c_2)$ and $x=pmsqrt{-c_1c_2}$ (in particular, $c_1c_2<0$). Differentiating gives the slopes are $-frac{c_i}{x}$ and so multiplies to $-1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Oops, yes that is indeed not given. Corrected.
    $endgroup$
    – user10354138
    Dec 9 '18 at 15:47


















1












$begingroup$

If the parabolae are conguent: Suppose the directrices are $y=c$ and $y=-c$. Then the equations of the parabolae are:
$$Gamma_pm:=(x^2+y^2=(ymp c)^2)=(x^2pm2yc-c^2=0)$$
They intersect at $(pm c,0)$ and it is easy to see their tangents have slope $pm 1$ there.



If the parabolae are not congruent: the directrices $y=c_i$, $i=1,2$ gives
$$
Gamma_i:=(x^2+2yc_i-c_i^2=0)
$$
so $y=frac12(c_1+c_2)$ and $x=pmsqrt{-c_1c_2}$ (in particular, $c_1c_2<0$). Differentiating gives the slopes are $-frac{c_i}{x}$ and so multiplies to $-1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Oops, yes that is indeed not given. Corrected.
    $endgroup$
    – user10354138
    Dec 9 '18 at 15:47
















1












1








1





$begingroup$

If the parabolae are conguent: Suppose the directrices are $y=c$ and $y=-c$. Then the equations of the parabolae are:
$$Gamma_pm:=(x^2+y^2=(ymp c)^2)=(x^2pm2yc-c^2=0)$$
They intersect at $(pm c,0)$ and it is easy to see their tangents have slope $pm 1$ there.



If the parabolae are not congruent: the directrices $y=c_i$, $i=1,2$ gives
$$
Gamma_i:=(x^2+2yc_i-c_i^2=0)
$$
so $y=frac12(c_1+c_2)$ and $x=pmsqrt{-c_1c_2}$ (in particular, $c_1c_2<0$). Differentiating gives the slopes are $-frac{c_i}{x}$ and so multiplies to $-1$.






share|cite|improve this answer











$endgroup$



If the parabolae are conguent: Suppose the directrices are $y=c$ and $y=-c$. Then the equations of the parabolae are:
$$Gamma_pm:=(x^2+y^2=(ymp c)^2)=(x^2pm2yc-c^2=0)$$
They intersect at $(pm c,0)$ and it is easy to see their tangents have slope $pm 1$ there.



If the parabolae are not congruent: the directrices $y=c_i$, $i=1,2$ gives
$$
Gamma_i:=(x^2+2yc_i-c_i^2=0)
$$
so $y=frac12(c_1+c_2)$ and $x=pmsqrt{-c_1c_2}$ (in particular, $c_1c_2<0$). Differentiating gives the slopes are $-frac{c_i}{x}$ and so multiplies to $-1$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 9 '18 at 15:46

























answered Dec 9 '18 at 10:02









user10354138user10354138

7,4322925




7,4322925












  • $begingroup$
    Oops, yes that is indeed not given. Corrected.
    $endgroup$
    – user10354138
    Dec 9 '18 at 15:47




















  • $begingroup$
    Oops, yes that is indeed not given. Corrected.
    $endgroup$
    – user10354138
    Dec 9 '18 at 15:47


















$begingroup$
Oops, yes that is indeed not given. Corrected.
$endgroup$
– user10354138
Dec 9 '18 at 15:47






$begingroup$
Oops, yes that is indeed not given. Corrected.
$endgroup$
– user10354138
Dec 9 '18 at 15:47




















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