Find the minimum sum of distances from a point on the circle with equation $x^2+y^2=16$ to the points...
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Find the minimum sum of distances from a point on the circle with equation $x^2+y^2=16$ to the points $A(-2,0)$ and $B(2,0)$.
Does this mean I have to find two distances, one from a point on the circle to point $A$ and one from a point on the circle to point $B$? Or am I supposed to find the distance between all $3$ points? If it's for all $3$, how do I go about doing this? thank you
geometry
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Find the minimum sum of distances from a point on the circle with equation $x^2+y^2=16$ to the points $A(-2,0)$ and $B(2,0)$.
Does this mean I have to find two distances, one from a point on the circle to point $A$ and one from a point on the circle to point $B$? Or am I supposed to find the distance between all $3$ points? If it's for all $3$, how do I go about doing this? thank you
geometry
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up vote
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favorite
up vote
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down vote
favorite
Find the minimum sum of distances from a point on the circle with equation $x^2+y^2=16$ to the points $A(-2,0)$ and $B(2,0)$.
Does this mean I have to find two distances, one from a point on the circle to point $A$ and one from a point on the circle to point $B$? Or am I supposed to find the distance between all $3$ points? If it's for all $3$, how do I go about doing this? thank you
geometry
Find the minimum sum of distances from a point on the circle with equation $x^2+y^2=16$ to the points $A(-2,0)$ and $B(2,0)$.
Does this mean I have to find two distances, one from a point on the circle to point $A$ and one from a point on the circle to point $B$? Or am I supposed to find the distance between all $3$ points? If it's for all $3$, how do I go about doing this? thank you
geometry
geometry
edited Nov 17 at 9:20
Tianlalu
2,844732
2,844732
asked Nov 17 at 9:03
user604369
235
235
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3 Answers
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You are supposed to minimize the sum of $overline{PA}$ and $overline{PB}$ where $P$ is your point on the circle.
You have to set up the system:
Minimize $sqrt{(x+2)^2+y^2}+sqrt{(x-2)^2+y^2}$ under the condition $x^2+y^2=16$
Hint: Use the circle condition to express $x$ by $y$ (or vice versa) in order to eliminate one variable.
The hint is a bit redundant. The expressions under $sqrt{}$ immediately simplify to expressions that only contain $x$ if $x^2+y^2=16$ is applied.
– drhab
Nov 17 at 9:19
Yes, I just suggested the application ;)
– Nodt Greenish
Nov 17 at 9:40
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0
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Your first sentence is correct. You just need to find two distances. Since you want the minimum, then it will have to be along the line from the point to the center of the circle. The sum of these two distances will then be minimum. Also, these distances will be the distance of the points from the origin minus the radius of the circle.
To be found is the minimum sum. Not the sum of two minimal terms.
– drhab
Nov 17 at 9:22
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Consider an ellipse with foci $(-2,0)$, and $(2,0).$
Let the sum of distances of a point $P$ to $(-2,0)$ and $(2,0)$ be $2a.$
Then $b^2= a^2 - 4$, where $b$ is the minor axis, and $a$ the major axis.
The locus of $P$:
$dfrac{x^2}{a^2} +dfrac{y^2}{b^2} =1.$
Minimize $a$ with the restriction:
$P$ is a point on
$x^2+y^2=16.$
$a$ is minimal when the $x=pm$ a, $y=0$. (Why?).
Hence $a^2=16$, $a= 4.$
Recall the sum of distances to to the foci is $2a$ ,
and finally $2a_{min}=8.$
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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
You are supposed to minimize the sum of $overline{PA}$ and $overline{PB}$ where $P$ is your point on the circle.
You have to set up the system:
Minimize $sqrt{(x+2)^2+y^2}+sqrt{(x-2)^2+y^2}$ under the condition $x^2+y^2=16$
Hint: Use the circle condition to express $x$ by $y$ (or vice versa) in order to eliminate one variable.
The hint is a bit redundant. The expressions under $sqrt{}$ immediately simplify to expressions that only contain $x$ if $x^2+y^2=16$ is applied.
– drhab
Nov 17 at 9:19
Yes, I just suggested the application ;)
– Nodt Greenish
Nov 17 at 9:40
add a comment |
up vote
1
down vote
You are supposed to minimize the sum of $overline{PA}$ and $overline{PB}$ where $P$ is your point on the circle.
You have to set up the system:
Minimize $sqrt{(x+2)^2+y^2}+sqrt{(x-2)^2+y^2}$ under the condition $x^2+y^2=16$
Hint: Use the circle condition to express $x$ by $y$ (or vice versa) in order to eliminate one variable.
The hint is a bit redundant. The expressions under $sqrt{}$ immediately simplify to expressions that only contain $x$ if $x^2+y^2=16$ is applied.
– drhab
Nov 17 at 9:19
Yes, I just suggested the application ;)
– Nodt Greenish
Nov 17 at 9:40
add a comment |
up vote
1
down vote
up vote
1
down vote
You are supposed to minimize the sum of $overline{PA}$ and $overline{PB}$ where $P$ is your point on the circle.
You have to set up the system:
Minimize $sqrt{(x+2)^2+y^2}+sqrt{(x-2)^2+y^2}$ under the condition $x^2+y^2=16$
Hint: Use the circle condition to express $x$ by $y$ (or vice versa) in order to eliminate one variable.
You are supposed to minimize the sum of $overline{PA}$ and $overline{PB}$ where $P$ is your point on the circle.
You have to set up the system:
Minimize $sqrt{(x+2)^2+y^2}+sqrt{(x-2)^2+y^2}$ under the condition $x^2+y^2=16$
Hint: Use the circle condition to express $x$ by $y$ (or vice versa) in order to eliminate one variable.
edited Nov 17 at 9:14
answered Nov 17 at 9:07
Nodt Greenish
30013
30013
The hint is a bit redundant. The expressions under $sqrt{}$ immediately simplify to expressions that only contain $x$ if $x^2+y^2=16$ is applied.
– drhab
Nov 17 at 9:19
Yes, I just suggested the application ;)
– Nodt Greenish
Nov 17 at 9:40
add a comment |
The hint is a bit redundant. The expressions under $sqrt{}$ immediately simplify to expressions that only contain $x$ if $x^2+y^2=16$ is applied.
– drhab
Nov 17 at 9:19
Yes, I just suggested the application ;)
– Nodt Greenish
Nov 17 at 9:40
The hint is a bit redundant. The expressions under $sqrt{}$ immediately simplify to expressions that only contain $x$ if $x^2+y^2=16$ is applied.
– drhab
Nov 17 at 9:19
The hint is a bit redundant. The expressions under $sqrt{}$ immediately simplify to expressions that only contain $x$ if $x^2+y^2=16$ is applied.
– drhab
Nov 17 at 9:19
Yes, I just suggested the application ;)
– Nodt Greenish
Nov 17 at 9:40
Yes, I just suggested the application ;)
– Nodt Greenish
Nov 17 at 9:40
add a comment |
up vote
0
down vote
Your first sentence is correct. You just need to find two distances. Since you want the minimum, then it will have to be along the line from the point to the center of the circle. The sum of these two distances will then be minimum. Also, these distances will be the distance of the points from the origin minus the radius of the circle.
To be found is the minimum sum. Not the sum of two minimal terms.
– drhab
Nov 17 at 9:22
add a comment |
up vote
0
down vote
Your first sentence is correct. You just need to find two distances. Since you want the minimum, then it will have to be along the line from the point to the center of the circle. The sum of these two distances will then be minimum. Also, these distances will be the distance of the points from the origin minus the radius of the circle.
To be found is the minimum sum. Not the sum of two minimal terms.
– drhab
Nov 17 at 9:22
add a comment |
up vote
0
down vote
up vote
0
down vote
Your first sentence is correct. You just need to find two distances. Since you want the minimum, then it will have to be along the line from the point to the center of the circle. The sum of these two distances will then be minimum. Also, these distances will be the distance of the points from the origin minus the radius of the circle.
Your first sentence is correct. You just need to find two distances. Since you want the minimum, then it will have to be along the line from the point to the center of the circle. The sum of these two distances will then be minimum. Also, these distances will be the distance of the points from the origin minus the radius of the circle.
answered Nov 17 at 9:08
Janus
538
538
To be found is the minimum sum. Not the sum of two minimal terms.
– drhab
Nov 17 at 9:22
add a comment |
To be found is the minimum sum. Not the sum of two minimal terms.
– drhab
Nov 17 at 9:22
To be found is the minimum sum. Not the sum of two minimal terms.
– drhab
Nov 17 at 9:22
To be found is the minimum sum. Not the sum of two minimal terms.
– drhab
Nov 17 at 9:22
add a comment |
up vote
0
down vote
Consider an ellipse with foci $(-2,0)$, and $(2,0).$
Let the sum of distances of a point $P$ to $(-2,0)$ and $(2,0)$ be $2a.$
Then $b^2= a^2 - 4$, where $b$ is the minor axis, and $a$ the major axis.
The locus of $P$:
$dfrac{x^2}{a^2} +dfrac{y^2}{b^2} =1.$
Minimize $a$ with the restriction:
$P$ is a point on
$x^2+y^2=16.$
$a$ is minimal when the $x=pm$ a, $y=0$. (Why?).
Hence $a^2=16$, $a= 4.$
Recall the sum of distances to to the foci is $2a$ ,
and finally $2a_{min}=8.$
add a comment |
up vote
0
down vote
Consider an ellipse with foci $(-2,0)$, and $(2,0).$
Let the sum of distances of a point $P$ to $(-2,0)$ and $(2,0)$ be $2a.$
Then $b^2= a^2 - 4$, where $b$ is the minor axis, and $a$ the major axis.
The locus of $P$:
$dfrac{x^2}{a^2} +dfrac{y^2}{b^2} =1.$
Minimize $a$ with the restriction:
$P$ is a point on
$x^2+y^2=16.$
$a$ is minimal when the $x=pm$ a, $y=0$. (Why?).
Hence $a^2=16$, $a= 4.$
Recall the sum of distances to to the foci is $2a$ ,
and finally $2a_{min}=8.$
add a comment |
up vote
0
down vote
up vote
0
down vote
Consider an ellipse with foci $(-2,0)$, and $(2,0).$
Let the sum of distances of a point $P$ to $(-2,0)$ and $(2,0)$ be $2a.$
Then $b^2= a^2 - 4$, where $b$ is the minor axis, and $a$ the major axis.
The locus of $P$:
$dfrac{x^2}{a^2} +dfrac{y^2}{b^2} =1.$
Minimize $a$ with the restriction:
$P$ is a point on
$x^2+y^2=16.$
$a$ is minimal when the $x=pm$ a, $y=0$. (Why?).
Hence $a^2=16$, $a= 4.$
Recall the sum of distances to to the foci is $2a$ ,
and finally $2a_{min}=8.$
Consider an ellipse with foci $(-2,0)$, and $(2,0).$
Let the sum of distances of a point $P$ to $(-2,0)$ and $(2,0)$ be $2a.$
Then $b^2= a^2 - 4$, where $b$ is the minor axis, and $a$ the major axis.
The locus of $P$:
$dfrac{x^2}{a^2} +dfrac{y^2}{b^2} =1.$
Minimize $a$ with the restriction:
$P$ is a point on
$x^2+y^2=16.$
$a$ is minimal when the $x=pm$ a, $y=0$. (Why?).
Hence $a^2=16$, $a= 4.$
Recall the sum of distances to to the foci is $2a$ ,
and finally $2a_{min}=8.$
answered Nov 17 at 10:35
Peter Szilas
10.2k2720
10.2k2720
add a comment |
add a comment |
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