How does one explicitly show that the distance from Focus_1 to the ellipse plus the distance from Focus_2 to...
$begingroup$
When deriving the equation for an ellipse, I have seen that one of the first steps is to recognize that for a point on the ellipse defined by y=0, the two distances from Focus_1 (call it D1) and Focus_2 (call it D2) to the ellipse add up to 2*a, where 'a' is the length from the origin to the ellipse at y = 0.
Logically, as long as you preface this by stating that "If there are two foci that are equidistant from the origin..." then this makes sense.
However, I am then confused when this equivalency (D1 + D2 = 2*a) is subsequently used for the remainder of the derivation. D1 changes its value as you move around the ellipse (i.e. when you're no longer at y = 0); D2, of course, changes its value as well.
Since D1 + D2 = 2a was only determined for the condition that y=0, why can I subsequently use this relationship for different y values on this ellipse. For all I know, D1 + D2 does not equal 2*a for any other y value. It seems like a logical inconsistency.
Am I thinking about this incorrectly?
geometry logic conic-sections
asked Jan 4 at 5:06
S.CramerS.Cramer
13618
$endgroup$
add a comment |
$begingroup$
When deriving the equation for an ellipse, I have seen that one of the first steps is to recognize that for a point on the ellipse defined by y=0, the two distances from Focus_1 (call it D1) and Focus_2 (call it D2) to the ellipse add up to 2*a, where 'a' is the length from the origin to the ellipse at y = 0.
Logically, as long as you preface this by stating that "If there are two foci that are equidistant from the origin..." then this makes sense.
However, I am then confused when this equivalency (D1 + D2 = 2*a) is subsequently used for the remainder of the derivation. D1 changes its value as you move around the ellipse (i.e. when you're no longer at y = 0); D2, of course, changes its value as well.
Since D1 + D2 = 2a was only determined for the condition that y=0, why can I subsequently use this relationship for different y values on this ellipse. For all I know, D1 + D2 does not equal 2*a for any other y value. It seems like a logical inconsistency.
Am I thinking about this incorrectly?
geometry logic conic-sections
asked Jan 4 at 5:06
S.CramerS.Cramer
13618
$endgroup$
add a comment |
$begingroup$
When deriving the equation for an ellipse, I have seen that one of the first steps is to recognize that for a point on the ellipse defined by y=0, the two distances from Focus_1 (call it D1) and Focus_2 (call it D2) to the ellipse add up to 2*a, where 'a' is the length from the origin to the ellipse at y = 0.
Logically, as long as you preface this by stating that "If there are two foci that are equidistant from the origin..." then this makes sense.
However, I am then confused when this equivalency (D1 + D2 = 2*a) is subsequently used for the remainder of the derivation. D1 changes its value as you move around the ellipse (i.e. when you're no longer at y = 0); D2, of course, changes its value as well.
Since D1 + D2 = 2a was only determined for the condition that y=0, why can I subsequently use this relationship for different y values on this ellipse. For all I know, D1 + D2 does not equal 2*a for any other y value. It seems like a logical inconsistency.
Am I thinking about this incorrectly?
geometry logic conic-sections
asked Jan 4 at 5:06
S.CramerS.Cramer
13618
$endgroup$
When deriving the equation for an ellipse, I have seen that one of the first steps is to recognize that for a point on the ellipse defined by y=0, the two distances from Focus_1 (call it D1) and Focus_2 (call it D2) to the ellipse add up to 2*a, where 'a' is the length from the origin to the ellipse at y = 0.
Logically, as long as you preface this by stating that "If there are two foci that are equidistant from the origin..." then this makes sense.
However, I am then confused when this equivalency (D1 + D2 = 2*a) is subsequently used for the remainder of the derivation. D1 changes its value as you move around the ellipse (i.e. when you're no longer at y = 0); D2, of course, changes its value as well.
Since D1 + D2 = 2a was only determined for the condition that y=0, why can I subsequently use this relationship for different y values on this ellipse. For all I know, D1 + D2 does not equal 2*a for any other y value. It seems like a logical inconsistency.
Am I thinking about this incorrectly?
geometry logic conic-sections
geometry logic conic-sections
asked Jan 4 at 5:06
S.CramerS.Cramer
13618
asked Jan 4 at 5:06
S.CramerS.Cramer
13618
asked Jan 4 at 5:06
S.CramerS.Cramer
13618
asked Jan 4 at 5:06
S.CramerS.Cramer
13618
asked Jan 4 at 5:06
S.CramerS.Cramer
13618
13618
add a comment |
add a comment |
2 Answers
2
active
oldest
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$begingroup$
The definition of the ellipse is the set of points whose total distance from the two fixed points called foci is a constant.
Since this constant at one instance is $2a$ it has to be $2a$ for every other point on the ellipse as well.
answered Jan 4 at 5:12
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
$endgroup$
$begingroup$
So, if I were to organize this derivation into a proof-like set up, I would do the following: 1) Assume the foci are equidistant from the origin 2) Assume the distance from each focus to the same point on the ellipse sum to a constant i.e. I have to state "if these two assumptions are true THEN..."
$endgroup$
– S.Cramer
Jan 4 at 5:14
1
$begingroup$
yes, it is much easier to put the origin at the center of the ellipse.
$endgroup$
– Mohammad Riazi-Kermani
Jan 4 at 5:16
add a comment |
$begingroup$
There extists some set of points $(x,y)$ shuch that the distance from this point to each focus is constant.
$d((c,0),(x,y)) + d((-c,0),(x,y)) = 2a\
sqrt{(x-c)^2 + y^2} + sqrt{(x+c)^2 + y^2} = 2a$
Now we have some algebra to cruch through...Square both sides
$(x-c)^2 + y^2 + (x+c)^2 + y^2 + 2sqrt{((x-c)^2 + y^2)((x+c)^2 + y^2)} = 4a^2$
Isolate the radical on one side and the terms not under the radical on the other.
$sqrt{((x-c)^2 + y^2)((x+c)^2 + y^2)} = 2a^2-x^2-y^2 -c^2$
square both sides again.
$(x^2 + y^2 +c^2- 2xc)(x^2 + y^2+ c^2 +2xc) = (2a^2 - x^2-y^2 - c^2)^2$
Multiply the two sides out.
$x^4 + y^4 +c^4 - 4x^2c^2 +2x^2y^2 +2x^2c^2 + 2y^2c^2= 4a^4+x^4+y^4+c^4 - 4a^2x^2-4a^2y^2 - 4a^2c^2+ 2x^2y^2 + 2x^2 c^2 + 2y^2c^2$
Subtract like terms from both sides.
$- 4x^2c^2 = 4a^4 - 4a^2x^2-4a^2y^2 - 4a^2c^2$
Move the x and y terms to one side and the contant terms to the other.
$4(a^2-c^2)x^2 + 4a^2y^2 = 4a^2(a^2-c^2)\
frac {x^2}{a^2} + frac {y^2}{a^2-c^2} = 1$
let $a^2-c^2 = b^2$
$frac {x^2}{a^2} + frac {y^2}{b^2} = 1$
answered Jan 4 at 5:31
Doug MDoug M
45.3k31954
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
The definition of the ellipse is the set of points whose total distance from the two fixed points called foci is a constant.
Since this constant at one instance is $2a$ it has to be $2a$ for every other point on the ellipse as well.
answered Jan 4 at 5:12
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
$endgroup$
$begingroup$
So, if I were to organize this derivation into a proof-like set up, I would do the following: 1) Assume the foci are equidistant from the origin 2) Assume the distance from each focus to the same point on the ellipse sum to a constant i.e. I have to state "if these two assumptions are true THEN..."
$endgroup$
– S.Cramer
Jan 4 at 5:14
1
$begingroup$
yes, it is much easier to put the origin at the center of the ellipse.
$endgroup$
– Mohammad Riazi-Kermani
Jan 4 at 5:16
add a comment |
$begingroup$
The definition of the ellipse is the set of points whose total distance from the two fixed points called foci is a constant.
Since this constant at one instance is $2a$ it has to be $2a$ for every other point on the ellipse as well.
answered Jan 4 at 5:12
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
$endgroup$
$begingroup$
So, if I were to organize this derivation into a proof-like set up, I would do the following: 1) Assume the foci are equidistant from the origin 2) Assume the distance from each focus to the same point on the ellipse sum to a constant i.e. I have to state "if these two assumptions are true THEN..."
$endgroup$
– S.Cramer
Jan 4 at 5:14
1
$begingroup$
yes, it is much easier to put the origin at the center of the ellipse.
$endgroup$
– Mohammad Riazi-Kermani
Jan 4 at 5:16
add a comment |
$begingroup$
The definition of the ellipse is the set of points whose total distance from the two fixed points called foci is a constant.
Since this constant at one instance is $2a$ it has to be $2a$ for every other point on the ellipse as well.
answered Jan 4 at 5:12
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
$endgroup$
The definition of the ellipse is the set of points whose total distance from the two fixed points called foci is a constant.
Since this constant at one instance is $2a$ it has to be $2a$ for every other point on the ellipse as well.
answered Jan 4 at 5:12
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
answered Jan 4 at 5:12
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
answered Jan 4 at 5:12
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
answered Jan 4 at 5:12
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.5k42061
41.5k42061
$begingroup$
So, if I were to organize this derivation into a proof-like set up, I would do the following: 1) Assume the foci are equidistant from the origin 2) Assume the distance from each focus to the same point on the ellipse sum to a constant i.e. I have to state "if these two assumptions are true THEN..."
$endgroup$
– S.Cramer
Jan 4 at 5:14
1
$begingroup$
yes, it is much easier to put the origin at the center of the ellipse.
$endgroup$
– Mohammad Riazi-Kermani
Jan 4 at 5:16
add a comment |
$begingroup$
So, if I were to organize this derivation into a proof-like set up, I would do the following: 1) Assume the foci are equidistant from the origin 2) Assume the distance from each focus to the same point on the ellipse sum to a constant i.e. I have to state "if these two assumptions are true THEN..."
$endgroup$
– S.Cramer
Jan 4 at 5:14
1
$begingroup$
yes, it is much easier to put the origin at the center of the ellipse.
$endgroup$
– Mohammad Riazi-Kermani
Jan 4 at 5:16
$begingroup$
So, if I were to organize this derivation into a proof-like set up, I would do the following: 1) Assume the foci are equidistant from the origin 2) Assume the distance from each focus to the same point on the ellipse sum to a constant i.e. I have to state "if these two assumptions are true THEN..."
$endgroup$
– S.Cramer
Jan 4 at 5:14
$begingroup$
So, if I were to organize this derivation into a proof-like set up, I would do the following: 1) Assume the foci are equidistant from the origin 2) Assume the distance from each focus to the same point on the ellipse sum to a constant i.e. I have to state "if these two assumptions are true THEN..."
$endgroup$
– S.Cramer
Jan 4 at 5:14
1
1
$begingroup$
yes, it is much easier to put the origin at the center of the ellipse.
$endgroup$
– Mohammad Riazi-Kermani
Jan 4 at 5:16
$begingroup$
yes, it is much easier to put the origin at the center of the ellipse.
$endgroup$
– Mohammad Riazi-Kermani
Jan 4 at 5:16
add a comment |
$begingroup$
There extists some set of points $(x,y)$ shuch that the distance from this point to each focus is constant.
$d((c,0),(x,y)) + d((-c,0),(x,y)) = 2a\
sqrt{(x-c)^2 + y^2} + sqrt{(x+c)^2 + y^2} = 2a$
Now we have some algebra to cruch through...Square both sides
$(x-c)^2 + y^2 + (x+c)^2 + y^2 + 2sqrt{((x-c)^2 + y^2)((x+c)^2 + y^2)} = 4a^2$
Isolate the radical on one side and the terms not under the radical on the other.
$sqrt{((x-c)^2 + y^2)((x+c)^2 + y^2)} = 2a^2-x^2-y^2 -c^2$
square both sides again.
$(x^2 + y^2 +c^2- 2xc)(x^2 + y^2+ c^2 +2xc) = (2a^2 - x^2-y^2 - c^2)^2$
Multiply the two sides out.
$x^4 + y^4 +c^4 - 4x^2c^2 +2x^2y^2 +2x^2c^2 + 2y^2c^2= 4a^4+x^4+y^4+c^4 - 4a^2x^2-4a^2y^2 - 4a^2c^2+ 2x^2y^2 + 2x^2 c^2 + 2y^2c^2$
Subtract like terms from both sides.
$- 4x^2c^2 = 4a^4 - 4a^2x^2-4a^2y^2 - 4a^2c^2$
Move the x and y terms to one side and the contant terms to the other.
$4(a^2-c^2)x^2 + 4a^2y^2 = 4a^2(a^2-c^2)\
frac {x^2}{a^2} + frac {y^2}{a^2-c^2} = 1$
let $a^2-c^2 = b^2$
$frac {x^2}{a^2} + frac {y^2}{b^2} = 1$
answered Jan 4 at 5:31
Doug MDoug M
45.3k31954
$endgroup$
add a comment |
$begingroup$
There extists some set of points $(x,y)$ shuch that the distance from this point to each focus is constant.
$d((c,0),(x,y)) + d((-c,0),(x,y)) = 2a\
sqrt{(x-c)^2 + y^2} + sqrt{(x+c)^2 + y^2} = 2a$
Now we have some algebra to cruch through...Square both sides
$(x-c)^2 + y^2 + (x+c)^2 + y^2 + 2sqrt{((x-c)^2 + y^2)((x+c)^2 + y^2)} = 4a^2$
Isolate the radical on one side and the terms not under the radical on the other.
$sqrt{((x-c)^2 + y^2)((x+c)^2 + y^2)} = 2a^2-x^2-y^2 -c^2$
square both sides again.
$(x^2 + y^2 +c^2- 2xc)(x^2 + y^2+ c^2 +2xc) = (2a^2 - x^2-y^2 - c^2)^2$
Multiply the two sides out.
$x^4 + y^4 +c^4 - 4x^2c^2 +2x^2y^2 +2x^2c^2 + 2y^2c^2= 4a^4+x^4+y^4+c^4 - 4a^2x^2-4a^2y^2 - 4a^2c^2+ 2x^2y^2 + 2x^2 c^2 + 2y^2c^2$
Subtract like terms from both sides.
$- 4x^2c^2 = 4a^4 - 4a^2x^2-4a^2y^2 - 4a^2c^2$
Move the x and y terms to one side and the contant terms to the other.
$4(a^2-c^2)x^2 + 4a^2y^2 = 4a^2(a^2-c^2)\
frac {x^2}{a^2} + frac {y^2}{a^2-c^2} = 1$
let $a^2-c^2 = b^2$
$frac {x^2}{a^2} + frac {y^2}{b^2} = 1$
answered Jan 4 at 5:31
Doug MDoug M
45.3k31954
$endgroup$
add a comment |
$begingroup$
There extists some set of points $(x,y)$ shuch that the distance from this point to each focus is constant.
$d((c,0),(x,y)) + d((-c,0),(x,y)) = 2a\
sqrt{(x-c)^2 + y^2} + sqrt{(x+c)^2 + y^2} = 2a$
Now we have some algebra to cruch through...Square both sides
$(x-c)^2 + y^2 + (x+c)^2 + y^2 + 2sqrt{((x-c)^2 + y^2)((x+c)^2 + y^2)} = 4a^2$
Isolate the radical on one side and the terms not under the radical on the other.
$sqrt{((x-c)^2 + y^2)((x+c)^2 + y^2)} = 2a^2-x^2-y^2 -c^2$
square both sides again.
$(x^2 + y^2 +c^2- 2xc)(x^2 + y^2+ c^2 +2xc) = (2a^2 - x^2-y^2 - c^2)^2$
Multiply the two sides out.
$x^4 + y^4 +c^4 - 4x^2c^2 +2x^2y^2 +2x^2c^2 + 2y^2c^2= 4a^4+x^4+y^4+c^4 - 4a^2x^2-4a^2y^2 - 4a^2c^2+ 2x^2y^2 + 2x^2 c^2 + 2y^2c^2$
Subtract like terms from both sides.
$- 4x^2c^2 = 4a^4 - 4a^2x^2-4a^2y^2 - 4a^2c^2$
Move the x and y terms to one side and the contant terms to the other.
$4(a^2-c^2)x^2 + 4a^2y^2 = 4a^2(a^2-c^2)\
frac {x^2}{a^2} + frac {y^2}{a^2-c^2} = 1$
let $a^2-c^2 = b^2$
$frac {x^2}{a^2} + frac {y^2}{b^2} = 1$
answered Jan 4 at 5:31
Doug MDoug M
45.3k31954
$endgroup$
There extists some set of points $(x,y)$ shuch that the distance from this point to each focus is constant.
$d((c,0),(x,y)) + d((-c,0),(x,y)) = 2a\
sqrt{(x-c)^2 + y^2} + sqrt{(x+c)^2 + y^2} = 2a$
Now we have some algebra to cruch through...Square both sides
$(x-c)^2 + y^2 + (x+c)^2 + y^2 + 2sqrt{((x-c)^2 + y^2)((x+c)^2 + y^2)} = 4a^2$
Isolate the radical on one side and the terms not under the radical on the other.
$sqrt{((x-c)^2 + y^2)((x+c)^2 + y^2)} = 2a^2-x^2-y^2 -c^2$
square both sides again.
$(x^2 + y^2 +c^2- 2xc)(x^2 + y^2+ c^2 +2xc) = (2a^2 - x^2-y^2 - c^2)^2$
Multiply the two sides out.
$x^4 + y^4 +c^4 - 4x^2c^2 +2x^2y^2 +2x^2c^2 + 2y^2c^2= 4a^4+x^4+y^4+c^4 - 4a^2x^2-4a^2y^2 - 4a^2c^2+ 2x^2y^2 + 2x^2 c^2 + 2y^2c^2$
Subtract like terms from both sides.
$- 4x^2c^2 = 4a^4 - 4a^2x^2-4a^2y^2 - 4a^2c^2$
Move the x and y terms to one side and the contant terms to the other.
$4(a^2-c^2)x^2 + 4a^2y^2 = 4a^2(a^2-c^2)\
frac {x^2}{a^2} + frac {y^2}{a^2-c^2} = 1$
let $a^2-c^2 = b^2$
$frac {x^2}{a^2} + frac {y^2}{b^2} = 1$
answered Jan 4 at 5:31
Doug MDoug M
45.3k31954
answered Jan 4 at 5:31
Doug MDoug M
45.3k31954
answered Jan 4 at 5:31
Doug MDoug M
45.3k31954
answered Jan 4 at 5:31
Doug MDoug M
45.3k31954
45.3k31954
add a comment |
add a comment |
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