Prove that the parabloas are mutually perpendicular.
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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
$endgroup$
|
show 2 more comments
$begingroup$
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
$endgroup$
$begingroup$
What do you mean by "axes ... in opposite direction"?
$endgroup$
– user10354138
Dec 9 '18 at 9:41
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@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
|
show 2 more comments
$begingroup$
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
$endgroup$
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
asked Dec 9 '18 at 9:13
Awe Kumar JhaAwe Kumar Jha
39613
39613
$begingroup$
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
|
show 2 more comments
$begingroup$
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
|
show 2 more comments
1 Answer
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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 $-1$.
$endgroup$
$begingroup$
Oops, yes that is indeed not given. Corrected.
$endgroup$
– user10354138
Dec 9 '18 at 15:47
add a comment |
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$begingroup$
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 $-1$.
$endgroup$
$begingroup$
Oops, yes that is indeed not given. Corrected.
$endgroup$
– user10354138
Dec 9 '18 at 15:47
add a comment |
$begingroup$
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 $-1$.
$endgroup$
$begingroup$
Oops, yes that is indeed not given. Corrected.
$endgroup$
– user10354138
Dec 9 '18 at 15:47
add a comment |
$begingroup$
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 $-1$.
$endgroup$
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 $-1$.
edited Dec 9 '18 at 15:46
answered Dec 9 '18 at 10:02
user10354138user10354138
7,3772925
7,3772925
$begingroup$
Oops, yes that is indeed not given. Corrected.
$endgroup$
– user10354138
Dec 9 '18 at 15:47
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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