Points $A$ and $B$ lie on the parabola $y=2x^2+4x-2,$such that the origin is the mid point of the line...
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Points $A$ and $B$ lie on the parabola $y=2x^2+4x-2,$such that the origin is the mid point of the line segment $AB$.Find the length of the line segment $AB$
$y=2(x^2+2x-1)=2(x+1)^2-4implies (y+4)=2(x+2)^2$ and let $x=t-2,y=2t^2-4$ be the parametric equation of the parabola.
I am stuck here.
conic-sections
asked Nov 26 at 12:38
user984325
14612
add a comment |
up vote
1
down vote
favorite
Points $A$ and $B$ lie on the parabola $y=2x^2+4x-2,$such that the origin is the mid point of the line segment $AB$.Find the length of the line segment $AB$
$y=2(x^2+2x-1)=2(x+1)^2-4implies (y+4)=2(x+2)^2$ and let $x=t-2,y=2t^2-4$ be the parametric equation of the parabola.
I am stuck here.
conic-sections
asked Nov 26 at 12:38
user984325
14612
Your parameterization is incorrect.
– amd
Nov 27 at 3:48
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Points $A$ and $B$ lie on the parabola $y=2x^2+4x-2,$such that the origin is the mid point of the line segment $AB$.Find the length of the line segment $AB$
$y=2(x^2+2x-1)=2(x+1)^2-4implies (y+4)=2(x+2)^2$ and let $x=t-2,y=2t^2-4$ be the parametric equation of the parabola.
I am stuck here.
conic-sections
asked Nov 26 at 12:38
user984325
14612
Points $A$ and $B$ lie on the parabola $y=2x^2+4x-2,$such that the origin is the mid point of the line segment $AB$.Find the length of the line segment $AB$
$y=2(x^2+2x-1)=2(x+1)^2-4implies (y+4)=2(x+2)^2$ and let $x=t-2,y=2t^2-4$ be the parametric equation of the parabola.
I am stuck here.
conic-sections
conic-sections
asked Nov 26 at 12:38
user984325
14612
asked Nov 26 at 12:38
user984325
14612
asked Nov 26 at 12:38
user984325
14612
asked Nov 26 at 12:38
user984325
14612
asked Nov 26 at 12:38
user984325
14612
14612
Your parameterization is incorrect.
– amd
Nov 27 at 3:48
add a comment |
Your parameterization is incorrect.
– amd
Nov 27 at 3:48
Your parameterization is incorrect.
– amd
Nov 27 at 3:48
Your parameterization is incorrect.
– amd
Nov 27 at 3:48
add a comment |
3 Answers
3
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up vote
1
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accepted
Let the points be $A(x_1,2x_1^2+4x_1-2)$ and $B(x_2,2x_2^2+4x_2-2)$. Then:
$$begin{cases}frac12(x_1+x_2)=0\ frac12(2x_1^2+4x_1-2+2x_2^2+4x_2-2)=0end{cases} Rightarrow (x_1,x_2)=(pm 1,mp 1).$$
Hence: $A(1,4)$, $B(-1,-4)$ and $AB=sqrt{(-1-1)^2+(-4-4)^2}=sqrt{68}$.
answered Nov 27 at 8:41
farruhota
18.5k2736
add a comment |
up vote
1
down vote
I’m pretty sure that there must be a more clever approach than the following, but one way to attack this problem is to use the fact that the midpoints of parallel chords all lie on a line parallel to the parabola’s axis. Therefore, the slope of $AB$ is equal to the slope of the tangent to the parabola at $x=0$. That gets you the equation of this line, from which you can find the coordinates of $A$ and $B$.
answered Nov 26 at 23:48
amd
28.9k21049
,If $A(u-2,2u^2-4),B(v-2,2v^2-4)$,then slope of $AB=2(u+v)$ and the slope of the tangent to the parabola at $x=0$ is $4$,equating these gives $u+v=2$ but origin is also the mid point of $AB$ which gives $u+v=4$,it is not possible.
– user984325
Nov 27 at 2:02
@user984325 And yet, the attached illustration shows a perfectly good solution to your problem. The origin is clearly the midpoint of $AB$. Your parameterization is incorrect. Try again.
– amd
Nov 27 at 3:44
add a comment |
up vote
0
down vote
What does the condition, that the mid point of $AB$ is the origin, say about the two points $A$ and $B$? How can you express one by the other? Then you can solve the equation for two unique points.
answered Nov 26 at 12:46
Gnampfissimo
18011
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Let the points be $A(x_1,2x_1^2+4x_1-2)$ and $B(x_2,2x_2^2+4x_2-2)$. Then:
$$begin{cases}frac12(x_1+x_2)=0\ frac12(2x_1^2+4x_1-2+2x_2^2+4x_2-2)=0end{cases} Rightarrow (x_1,x_2)=(pm 1,mp 1).$$
Hence: $A(1,4)$, $B(-1,-4)$ and $AB=sqrt{(-1-1)^2+(-4-4)^2}=sqrt{68}$.
answered Nov 27 at 8:41
farruhota
18.5k2736
add a comment |
up vote
1
down vote
accepted
Let the points be $A(x_1,2x_1^2+4x_1-2)$ and $B(x_2,2x_2^2+4x_2-2)$. Then:
$$begin{cases}frac12(x_1+x_2)=0\ frac12(2x_1^2+4x_1-2+2x_2^2+4x_2-2)=0end{cases} Rightarrow (x_1,x_2)=(pm 1,mp 1).$$
Hence: $A(1,4)$, $B(-1,-4)$ and $AB=sqrt{(-1-1)^2+(-4-4)^2}=sqrt{68}$.
answered Nov 27 at 8:41
farruhota
18.5k2736
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let the points be $A(x_1,2x_1^2+4x_1-2)$ and $B(x_2,2x_2^2+4x_2-2)$. Then:
$$begin{cases}frac12(x_1+x_2)=0\ frac12(2x_1^2+4x_1-2+2x_2^2+4x_2-2)=0end{cases} Rightarrow (x_1,x_2)=(pm 1,mp 1).$$
Hence: $A(1,4)$, $B(-1,-4)$ and $AB=sqrt{(-1-1)^2+(-4-4)^2}=sqrt{68}$.
answered Nov 27 at 8:41
farruhota
18.5k2736
Let the points be $A(x_1,2x_1^2+4x_1-2)$ and $B(x_2,2x_2^2+4x_2-2)$. Then:
$$begin{cases}frac12(x_1+x_2)=0\ frac12(2x_1^2+4x_1-2+2x_2^2+4x_2-2)=0end{cases} Rightarrow (x_1,x_2)=(pm 1,mp 1).$$
Hence: $A(1,4)$, $B(-1,-4)$ and $AB=sqrt{(-1-1)^2+(-4-4)^2}=sqrt{68}$.
answered Nov 27 at 8:41
farruhota
18.5k2736
answered Nov 27 at 8:41
farruhota
18.5k2736
answered Nov 27 at 8:41
farruhota
18.5k2736
answered Nov 27 at 8:41
farruhota
18.5k2736
18.5k2736
add a comment |
add a comment |
up vote
1
down vote
I’m pretty sure that there must be a more clever approach than the following, but one way to attack this problem is to use the fact that the midpoints of parallel chords all lie on a line parallel to the parabola’s axis. Therefore, the slope of $AB$ is equal to the slope of the tangent to the parabola at $x=0$. That gets you the equation of this line, from which you can find the coordinates of $A$ and $B$.
answered Nov 26 at 23:48
amd
28.9k21049
,If $A(u-2,2u^2-4),B(v-2,2v^2-4)$,then slope of $AB=2(u+v)$ and the slope of the tangent to the parabola at $x=0$ is $4$,equating these gives $u+v=2$ but origin is also the mid point of $AB$ which gives $u+v=4$,it is not possible.
– user984325
Nov 27 at 2:02
@user984325 And yet, the attached illustration shows a perfectly good solution to your problem. The origin is clearly the midpoint of $AB$. Your parameterization is incorrect. Try again.
– amd
Nov 27 at 3:44
add a comment |
up vote
1
down vote
I’m pretty sure that there must be a more clever approach than the following, but one way to attack this problem is to use the fact that the midpoints of parallel chords all lie on a line parallel to the parabola’s axis. Therefore, the slope of $AB$ is equal to the slope of the tangent to the parabola at $x=0$. That gets you the equation of this line, from which you can find the coordinates of $A$ and $B$.
answered Nov 26 at 23:48
amd
28.9k21049
,If $A(u-2,2u^2-4),B(v-2,2v^2-4)$,then slope of $AB=2(u+v)$ and the slope of the tangent to the parabola at $x=0$ is $4$,equating these gives $u+v=2$ but origin is also the mid point of $AB$ which gives $u+v=4$,it is not possible.
– user984325
Nov 27 at 2:02
@user984325 And yet, the attached illustration shows a perfectly good solution to your problem. The origin is clearly the midpoint of $AB$. Your parameterization is incorrect. Try again.
– amd
Nov 27 at 3:44
add a comment |
up vote
1
down vote
up vote
1
down vote
I’m pretty sure that there must be a more clever approach than the following, but one way to attack this problem is to use the fact that the midpoints of parallel chords all lie on a line parallel to the parabola’s axis. Therefore, the slope of $AB$ is equal to the slope of the tangent to the parabola at $x=0$. That gets you the equation of this line, from which you can find the coordinates of $A$ and $B$.
answered Nov 26 at 23:48
amd
28.9k21049
I’m pretty sure that there must be a more clever approach than the following, but one way to attack this problem is to use the fact that the midpoints of parallel chords all lie on a line parallel to the parabola’s axis. Therefore, the slope of $AB$ is equal to the slope of the tangent to the parabola at $x=0$. That gets you the equation of this line, from which you can find the coordinates of $A$ and $B$.
answered Nov 26 at 23:48
amd
28.9k21049
edited Dec 1 at 0:31
answered Nov 26 at 23:48
amd
28.9k21049
answered Nov 26 at 23:48
amd
28.9k21049
answered Nov 26 at 23:48
amd
28.9k21049
28.9k21049
,If $A(u-2,2u^2-4),B(v-2,2v^2-4)$,then slope of $AB=2(u+v)$ and the slope of the tangent to the parabola at $x=0$ is $4$,equating these gives $u+v=2$ but origin is also the mid point of $AB$ which gives $u+v=4$,it is not possible.
– user984325
Nov 27 at 2:02
@user984325 And yet, the attached illustration shows a perfectly good solution to your problem. The origin is clearly the midpoint of $AB$. Your parameterization is incorrect. Try again.
– amd
Nov 27 at 3:44
add a comment |
,If $A(u-2,2u^2-4),B(v-2,2v^2-4)$,then slope of $AB=2(u+v)$ and the slope of the tangent to the parabola at $x=0$ is $4$,equating these gives $u+v=2$ but origin is also the mid point of $AB$ which gives $u+v=4$,it is not possible.
– user984325
Nov 27 at 2:02
@user984325 And yet, the attached illustration shows a perfectly good solution to your problem. The origin is clearly the midpoint of $AB$. Your parameterization is incorrect. Try again.
– amd
Nov 27 at 3:44
,If $A(u-2,2u^2-4),B(v-2,2v^2-4)$,then slope of $AB=2(u+v)$ and the slope of the tangent to the parabola at $x=0$ is $4$,equating these gives $u+v=2$ but origin is also the mid point of $AB$ which gives $u+v=4$,it is not possible.
– user984325
Nov 27 at 2:02
,If $A(u-2,2u^2-4),B(v-2,2v^2-4)$,then slope of $AB=2(u+v)$ and the slope of the tangent to the parabola at $x=0$ is $4$,equating these gives $u+v=2$ but origin is also the mid point of $AB$ which gives $u+v=4$,it is not possible.
– user984325
Nov 27 at 2:02
@user984325 And yet, the attached illustration shows a perfectly good solution to your problem. The origin is clearly the midpoint of $AB$. Your parameterization is incorrect. Try again.
– amd
Nov 27 at 3:44
@user984325 And yet, the attached illustration shows a perfectly good solution to your problem. The origin is clearly the midpoint of $AB$. Your parameterization is incorrect. Try again.
– amd
Nov 27 at 3:44
add a comment |
up vote
0
down vote
What does the condition, that the mid point of $AB$ is the origin, say about the two points $A$ and $B$? How can you express one by the other? Then you can solve the equation for two unique points.
answered Nov 26 at 12:46
Gnampfissimo
18011
add a comment |
up vote
0
down vote
What does the condition, that the mid point of $AB$ is the origin, say about the two points $A$ and $B$? How can you express one by the other? Then you can solve the equation for two unique points.
answered Nov 26 at 12:46
Gnampfissimo
18011
add a comment |
up vote
0
down vote
up vote
0
down vote
What does the condition, that the mid point of $AB$ is the origin, say about the two points $A$ and $B$? How can you express one by the other? Then you can solve the equation for two unique points.
answered Nov 26 at 12:46
Gnampfissimo
18011
What does the condition, that the mid point of $AB$ is the origin, say about the two points $A$ and $B$? How can you express one by the other? Then you can solve the equation for two unique points.
answered Nov 26 at 12:46
Gnampfissimo
18011
answered Nov 26 at 12:46
Gnampfissimo
18011
answered Nov 26 at 12:46
Gnampfissimo
18011
answered Nov 26 at 12:46
Gnampfissimo
18011
18011
add a comment |
add a comment |
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Your parameterization is incorrect.
– amd
Nov 27 at 3:48