Mixing
How to show $|F_1-v|+|F_2-v|=c$ for an ellipse with foci $F_1, F_2$
$begingroup$
I am asked to show the sum of the distances between the two foci on the ellipse and any point, $v$, on the ellipse is independent of one’s choice of $v$.
Starting with the standard equation of an ellipse: $$frac{x^2}{p^2}+frac{y^2}{q^2}=1$$ and a parametrization of it: $$gamma(t)=left(pcos(t), qsin(t)right)spacespace text{for}space tin[0,2pi]$$
The foci of the ellipse are $F_1=(ep,0)$ and $F_2=(-ep,0)$ where $e=sqrt{1-frac{q^2}{p^2}}$. My attempt to show the sum is constant is the following:
$$|v-F_1|=sqrt{(pcos(t)-ep)^2+(qsin(t)-0)^2}=sqrt{p^2cos^2(t)-2p^2ecos(t)+e^2p^2+q^2sin^2(t)}$$ $$|v-F_2|=sqrt{(pcos(t)+ep)^2+(qsin(t)-0)^2}=sqrt{p^2cos^2(t)+2p^2ecos(t)+e^2p^2+q^2sin^2(t)}$$
I am stuck here trying to show the sum of these two expressions is independent of $t$.
I would be very grateful for hints and not full solutions. Thanks in advance!
analytic-geometry conic-sections
edited Jan 10 at 2:45
darij grinberg
10.8k33062
asked Jan 10 at 1:24
livingtolearn-learningtolivelivingtolearn-learningtolive
16010
$endgroup$
add a comment |
$begingroup$
I am asked to show the sum of the distances between the two foci on the ellipse and any point, $v$, on the ellipse is independent of one’s choice of $v$.
Starting with the standard equation of an ellipse: $$frac{x^2}{p^2}+frac{y^2}{q^2}=1$$ and a parametrization of it: $$gamma(t)=left(pcos(t), qsin(t)right)spacespace text{for}space tin[0,2pi]$$
The foci of the ellipse are $F_1=(ep,0)$ and $F_2=(-ep,0)$ where $e=sqrt{1-frac{q^2}{p^2}}$. My attempt to show the sum is constant is the following:
$$|v-F_1|=sqrt{(pcos(t)-ep)^2+(qsin(t)-0)^2}=sqrt{p^2cos^2(t)-2p^2ecos(t)+e^2p^2+q^2sin^2(t)}$$ $$|v-F_2|=sqrt{(pcos(t)+ep)^2+(qsin(t)-0)^2}=sqrt{p^2cos^2(t)+2p^2ecos(t)+e^2p^2+q^2sin^2(t)}$$
I am stuck here trying to show the sum of these two expressions is independent of $t$.
I would be very grateful for hints and not full solutions. Thanks in advance!
analytic-geometry conic-sections
edited Jan 10 at 2:45
darij grinberg
10.8k33062
asked Jan 10 at 1:24
livingtolearn-learningtolivelivingtolearn-learningtolive
16010
$endgroup$
1
$begingroup$
Solve your definition of $e$ for $q^2$, and substitute that into the expressions for the distances.
$endgroup$
– Blue
Jan 10 at 1:39
1
$begingroup$
Thanks, @Blue! It took me awhile to see the nice factorization, but I finally did. Thanks again!
$endgroup$
– livingtolearn-learningtolive
Jan 10 at 3:23
$begingroup$
You should post your solution as an answer so that we can upvote it (and so that the question gets removed from the Unanswered queue).
$endgroup$
– Blue
Jan 10 at 3:25
1
$begingroup$
Done @blue. Thanks for you help again.
$endgroup$
– livingtolearn-learningtolive
Jan 10 at 4:01
add a comment |
$begingroup$
I am asked to show the sum of the distances between the two foci on the ellipse and any point, $v$, on the ellipse is independent of one’s choice of $v$.
Starting with the standard equation of an ellipse: $$frac{x^2}{p^2}+frac{y^2}{q^2}=1$$ and a parametrization of it: $$gamma(t)=left(pcos(t), qsin(t)right)spacespace text{for}space tin[0,2pi]$$
The foci of the ellipse are $F_1=(ep,0)$ and $F_2=(-ep,0)$ where $e=sqrt{1-frac{q^2}{p^2}}$. My attempt to show the sum is constant is the following:
$$|v-F_1|=sqrt{(pcos(t)-ep)^2+(qsin(t)-0)^2}=sqrt{p^2cos^2(t)-2p^2ecos(t)+e^2p^2+q^2sin^2(t)}$$ $$|v-F_2|=sqrt{(pcos(t)+ep)^2+(qsin(t)-0)^2}=sqrt{p^2cos^2(t)+2p^2ecos(t)+e^2p^2+q^2sin^2(t)}$$
I am stuck here trying to show the sum of these two expressions is independent of $t$.
I would be very grateful for hints and not full solutions. Thanks in advance!
analytic-geometry conic-sections
edited Jan 10 at 2:45
darij grinberg
10.8k33062
asked Jan 10 at 1:24
livingtolearn-learningtolivelivingtolearn-learningtolive
16010
$endgroup$
I am asked to show the sum of the distances between the two foci on the ellipse and any point, $v$, on the ellipse is independent of one’s choice of $v$.
Starting with the standard equation of an ellipse: $$frac{x^2}{p^2}+frac{y^2}{q^2}=1$$ and a parametrization of it: $$gamma(t)=left(pcos(t), qsin(t)right)spacespace text{for}space tin[0,2pi]$$
The foci of the ellipse are $F_1=(ep,0)$ and $F_2=(-ep,0)$ where $e=sqrt{1-frac{q^2}{p^2}}$. My attempt to show the sum is constant is the following:
$$|v-F_1|=sqrt{(pcos(t)-ep)^2+(qsin(t)-0)^2}=sqrt{p^2cos^2(t)-2p^2ecos(t)+e^2p^2+q^2sin^2(t)}$$ $$|v-F_2|=sqrt{(pcos(t)+ep)^2+(qsin(t)-0)^2}=sqrt{p^2cos^2(t)+2p^2ecos(t)+e^2p^2+q^2sin^2(t)}$$
I am stuck here trying to show the sum of these two expressions is independent of $t$.
I would be very grateful for hints and not full solutions. Thanks in advance!
analytic-geometry conic-sections
analytic-geometry conic-sections
edited Jan 10 at 2:45
darij grinberg
10.8k33062
asked Jan 10 at 1:24
livingtolearn-learningtolivelivingtolearn-learningtolive
16010
edited Jan 10 at 2:45
darij grinberg
10.8k33062
asked Jan 10 at 1:24
livingtolearn-learningtolivelivingtolearn-learningtolive
16010
edited Jan 10 at 2:45
darij grinberg
10.8k33062
edited Jan 10 at 2:45
darij grinberg
10.8k33062
edited Jan 10 at 2:45
darij grinberg
10.8k33062
10.8k33062
asked Jan 10 at 1:24
livingtolearn-learningtolivelivingtolearn-learningtolive
16010
asked Jan 10 at 1:24
livingtolearn-learningtolivelivingtolearn-learningtolive
16010
asked Jan 10 at 1:24
livingtolearn-learningtolivelivingtolearn-learningtolive
16010
16010
1
$begingroup$
Solve your definition of $e$ for $q^2$, and substitute that into the expressions for the distances.
$endgroup$
– Blue
Jan 10 at 1:39
1
$begingroup$
Thanks, @Blue! It took me awhile to see the nice factorization, but I finally did. Thanks again!
$endgroup$
– livingtolearn-learningtolive
Jan 10 at 3:23
$begingroup$
You should post your solution as an answer so that we can upvote it (and so that the question gets removed from the Unanswered queue).
$endgroup$
– Blue
Jan 10 at 3:25
1
$begingroup$
Done @blue. Thanks for you help again.
$endgroup$
– livingtolearn-learningtolive
Jan 10 at 4:01
add a comment |
1
$begingroup$
Solve your definition of $e$ for $q^2$, and substitute that into the expressions for the distances.
$endgroup$
– Blue
Jan 10 at 1:39
1
$begingroup$
Thanks, @Blue! It took me awhile to see the nice factorization, but I finally did. Thanks again!
$endgroup$
– livingtolearn-learningtolive
Jan 10 at 3:23
$begingroup$
You should post your solution as an answer so that we can upvote it (and so that the question gets removed from the Unanswered queue).
$endgroup$
– Blue
Jan 10 at 3:25
1
$begingroup$
Done @blue. Thanks for you help again.
$endgroup$
– livingtolearn-learningtolive
Jan 10 at 4:01
1
1
$begingroup$
Solve your definition of $e$ for $q^2$, and substitute that into the expressions for the distances.
$endgroup$
– Blue
Jan 10 at 1:39
$begingroup$
Solve your definition of $e$ for $q^2$, and substitute that into the expressions for the distances.
$endgroup$
– Blue
Jan 10 at 1:39
1
1
$begingroup$
Thanks, @Blue! It took me awhile to see the nice factorization, but I finally did. Thanks again!
$endgroup$
– livingtolearn-learningtolive
Jan 10 at 3:23
$begingroup$
Thanks, @Blue! It took me awhile to see the nice factorization, but I finally did. Thanks again!
$endgroup$
– livingtolearn-learningtolive
Jan 10 at 3:23
$begingroup$
You should post your solution as an answer so that we can upvote it (and so that the question gets removed from the Unanswered queue).
$endgroup$
– Blue
Jan 10 at 3:25
$begingroup$
You should post your solution as an answer so that we can upvote it (and so that the question gets removed from the Unanswered queue).
$endgroup$
– Blue
Jan 10 at 3:25
1
1
$begingroup$
Done @blue. Thanks for you help again.
$endgroup$
– livingtolearn-learningtolive
Jan 10 at 4:01
$begingroup$
Done @blue. Thanks for you help again.
$endgroup$
– livingtolearn-learningtolive
Jan 10 at 4:01
add a comment |
1 Answer
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$begingroup$
Per Blue’s request, I will finish my answer outlined above...
$$|v-F_1|=sqrt{(pcos(t)-ep)^2+(qsin(t)-0)^2}=sqrt{p^2cos^2(t)-2p^2ecos(t)+e^2p^2+q^2sin^2(t)}$$ $$|v-F_2|=sqrt{(pcos(t)+ep)^2+(qsin(t)-0)^2}=sqrt{p^2cos^2(t)+2p^2ecos(t)+e^2p^2+q^2sin^2(t)}$$
Use the change of variable $e=sqrt{1-frac{q^2}{p^2}}implies q^2=p^2-p^2e^2implies$ begin{equation} label{eq1}
begin{split}
|v-F_1|& = sqrt{p^2cos^2(t)-2p^2ecos(t)+e^2p^2+(p^2-p^2e^2)sin^2(t)} \
& = sqrt{p^2cos^2(t)-2p^2ecos(t)+e^2p^2+p^2sin^2(t)-p^2e^2sin^2(t)}\
&=sqrt{p^2-2p^2ecos(t)+e^2p^2-p^2e^2sin^2(t)} \
& =sqrt{p^2-2p^2ecos(t)+p^2e^2cos^2(t)}\
&=sqrt{p^2left(1-2ecos(t)+e^2cos^2(t)right)} \
&=sqrt{p^2left(1-ecos(t)right)^2} \
&=p(1-ecos(t))
end{split}
end{equation}
A similar lengthy calculation shows that: $$|v-F_2|=p(1+ecos(t))$$
Hence $|v-F_1|+|v-F_2|=p(1-ecos(t))+p(1+ecos(t))=2p$
Thus we have shown that the sum of the distances between both foci and a point, $v$, on the ellipse is constant namely $2p$.
answered Jan 10 at 4:00
livingtolearn-learningtolivelivingtolearn-learningtolive
16010
$endgroup$
1
$begingroup$
+1. Note, however, that $sqrt{x^2}=|x|$, so you really should say a word or two about why you can write simply (and correctly) $sqrt{p^2(1-ecos t)^2} = p(1-ecos t)$. (Oh, and be sure to "accept" your answer.)
$endgroup$
– Blue
Jan 10 at 4:07
add a comment |
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$begingroup$
Per Blue’s request, I will finish my answer outlined above...
$$|v-F_1|=sqrt{(pcos(t)-ep)^2+(qsin(t)-0)^2}=sqrt{p^2cos^2(t)-2p^2ecos(t)+e^2p^2+q^2sin^2(t)}$$ $$|v-F_2|=sqrt{(pcos(t)+ep)^2+(qsin(t)-0)^2}=sqrt{p^2cos^2(t)+2p^2ecos(t)+e^2p^2+q^2sin^2(t)}$$
Use the change of variable $e=sqrt{1-frac{q^2}{p^2}}implies q^2=p^2-p^2e^2implies$ begin{equation} label{eq1}
begin{split}
|v-F_1|& = sqrt{p^2cos^2(t)-2p^2ecos(t)+e^2p^2+(p^2-p^2e^2)sin^2(t)} \
& = sqrt{p^2cos^2(t)-2p^2ecos(t)+e^2p^2+p^2sin^2(t)-p^2e^2sin^2(t)}\
&=sqrt{p^2-2p^2ecos(t)+e^2p^2-p^2e^2sin^2(t)} \
& =sqrt{p^2-2p^2ecos(t)+p^2e^2cos^2(t)}\
&=sqrt{p^2left(1-2ecos(t)+e^2cos^2(t)right)} \
&=sqrt{p^2left(1-ecos(t)right)^2} \
&=p(1-ecos(t))
end{split}
end{equation}
A similar lengthy calculation shows that: $$|v-F_2|=p(1+ecos(t))$$
Hence $|v-F_1|+|v-F_2|=p(1-ecos(t))+p(1+ecos(t))=2p$
Thus we have shown that the sum of the distances between both foci and a point, $v$, on the ellipse is constant namely $2p$.
answered Jan 10 at 4:00
livingtolearn-learningtolivelivingtolearn-learningtolive
16010
$endgroup$
1
$begingroup$
+1. Note, however, that $sqrt{x^2}=|x|$, so you really should say a word or two about why you can write simply (and correctly) $sqrt{p^2(1-ecos t)^2} = p(1-ecos t)$. (Oh, and be sure to "accept" your answer.)
$endgroup$
– Blue
Jan 10 at 4:07
add a comment |
$begingroup$
Per Blue’s request, I will finish my answer outlined above...
$$|v-F_1|=sqrt{(pcos(t)-ep)^2+(qsin(t)-0)^2}=sqrt{p^2cos^2(t)-2p^2ecos(t)+e^2p^2+q^2sin^2(t)}$$ $$|v-F_2|=sqrt{(pcos(t)+ep)^2+(qsin(t)-0)^2}=sqrt{p^2cos^2(t)+2p^2ecos(t)+e^2p^2+q^2sin^2(t)}$$
Use the change of variable $e=sqrt{1-frac{q^2}{p^2}}implies q^2=p^2-p^2e^2implies$ begin{equation} label{eq1}
begin{split}
|v-F_1|& = sqrt{p^2cos^2(t)-2p^2ecos(t)+e^2p^2+(p^2-p^2e^2)sin^2(t)} \
& = sqrt{p^2cos^2(t)-2p^2ecos(t)+e^2p^2+p^2sin^2(t)-p^2e^2sin^2(t)}\
&=sqrt{p^2-2p^2ecos(t)+e^2p^2-p^2e^2sin^2(t)} \
& =sqrt{p^2-2p^2ecos(t)+p^2e^2cos^2(t)}\
&=sqrt{p^2left(1-2ecos(t)+e^2cos^2(t)right)} \
&=sqrt{p^2left(1-ecos(t)right)^2} \
&=p(1-ecos(t))
end{split}
end{equation}
A similar lengthy calculation shows that: $$|v-F_2|=p(1+ecos(t))$$
Hence $|v-F_1|+|v-F_2|=p(1-ecos(t))+p(1+ecos(t))=2p$
Thus we have shown that the sum of the distances between both foci and a point, $v$, on the ellipse is constant namely $2p$.
answered Jan 10 at 4:00
livingtolearn-learningtolivelivingtolearn-learningtolive
16010
$endgroup$
1
$begingroup$
+1. Note, however, that $sqrt{x^2}=|x|$, so you really should say a word or two about why you can write simply (and correctly) $sqrt{p^2(1-ecos t)^2} = p(1-ecos t)$. (Oh, and be sure to "accept" your answer.)
$endgroup$
– Blue
Jan 10 at 4:07
add a comment |
$begingroup$
Per Blue’s request, I will finish my answer outlined above...
$$|v-F_1|=sqrt{(pcos(t)-ep)^2+(qsin(t)-0)^2}=sqrt{p^2cos^2(t)-2p^2ecos(t)+e^2p^2+q^2sin^2(t)}$$ $$|v-F_2|=sqrt{(pcos(t)+ep)^2+(qsin(t)-0)^2}=sqrt{p^2cos^2(t)+2p^2ecos(t)+e^2p^2+q^2sin^2(t)}$$
Use the change of variable $e=sqrt{1-frac{q^2}{p^2}}implies q^2=p^2-p^2e^2implies$ begin{equation} label{eq1}
begin{split}
|v-F_1|& = sqrt{p^2cos^2(t)-2p^2ecos(t)+e^2p^2+(p^2-p^2e^2)sin^2(t)} \
& = sqrt{p^2cos^2(t)-2p^2ecos(t)+e^2p^2+p^2sin^2(t)-p^2e^2sin^2(t)}\
&=sqrt{p^2-2p^2ecos(t)+e^2p^2-p^2e^2sin^2(t)} \
& =sqrt{p^2-2p^2ecos(t)+p^2e^2cos^2(t)}\
&=sqrt{p^2left(1-2ecos(t)+e^2cos^2(t)right)} \
&=sqrt{p^2left(1-ecos(t)right)^2} \
&=p(1-ecos(t))
end{split}
end{equation}
A similar lengthy calculation shows that: $$|v-F_2|=p(1+ecos(t))$$
Hence $|v-F_1|+|v-F_2|=p(1-ecos(t))+p(1+ecos(t))=2p$
Thus we have shown that the sum of the distances between both foci and a point, $v$, on the ellipse is constant namely $2p$.
answered Jan 10 at 4:00
livingtolearn-learningtolivelivingtolearn-learningtolive
16010
$endgroup$
Per Blue’s request, I will finish my answer outlined above...
$$|v-F_1|=sqrt{(pcos(t)-ep)^2+(qsin(t)-0)^2}=sqrt{p^2cos^2(t)-2p^2ecos(t)+e^2p^2+q^2sin^2(t)}$$ $$|v-F_2|=sqrt{(pcos(t)+ep)^2+(qsin(t)-0)^2}=sqrt{p^2cos^2(t)+2p^2ecos(t)+e^2p^2+q^2sin^2(t)}$$
Use the change of variable $e=sqrt{1-frac{q^2}{p^2}}implies q^2=p^2-p^2e^2implies$ begin{equation} label{eq1}
begin{split}
|v-F_1|& = sqrt{p^2cos^2(t)-2p^2ecos(t)+e^2p^2+(p^2-p^2e^2)sin^2(t)} \
& = sqrt{p^2cos^2(t)-2p^2ecos(t)+e^2p^2+p^2sin^2(t)-p^2e^2sin^2(t)}\
&=sqrt{p^2-2p^2ecos(t)+e^2p^2-p^2e^2sin^2(t)} \
& =sqrt{p^2-2p^2ecos(t)+p^2e^2cos^2(t)}\
&=sqrt{p^2left(1-2ecos(t)+e^2cos^2(t)right)} \
&=sqrt{p^2left(1-ecos(t)right)^2} \
&=p(1-ecos(t))
end{split}
end{equation}
A similar lengthy calculation shows that: $$|v-F_2|=p(1+ecos(t))$$
Hence $|v-F_1|+|v-F_2|=p(1-ecos(t))+p(1+ecos(t))=2p$
Thus we have shown that the sum of the distances between both foci and a point, $v$, on the ellipse is constant namely $2p$.
answered Jan 10 at 4:00
livingtolearn-learningtolivelivingtolearn-learningtolive
16010
answered Jan 10 at 4:00
livingtolearn-learningtolivelivingtolearn-learningtolive
16010
answered Jan 10 at 4:00
livingtolearn-learningtolivelivingtolearn-learningtolive
16010
answered Jan 10 at 4:00
livingtolearn-learningtolivelivingtolearn-learningtolive
16010
16010
1
$begingroup$
+1. Note, however, that $sqrt{x^2}=|x|$, so you really should say a word or two about why you can write simply (and correctly) $sqrt{p^2(1-ecos t)^2} = p(1-ecos t)$. (Oh, and be sure to "accept" your answer.)
$endgroup$
– Blue
Jan 10 at 4:07
add a comment |
1
$begingroup$
+1. Note, however, that $sqrt{x^2}=|x|$, so you really should say a word or two about why you can write simply (and correctly) $sqrt{p^2(1-ecos t)^2} = p(1-ecos t)$. (Oh, and be sure to "accept" your answer.)
$endgroup$
– Blue
Jan 10 at 4:07
1
1
$begingroup$
+1. Note, however, that $sqrt{x^2}=|x|$, so you really should say a word or two about why you can write simply (and correctly) $sqrt{p^2(1-ecos t)^2} = p(1-ecos t)$. (Oh, and be sure to "accept" your answer.)
$endgroup$
– Blue
Jan 10 at 4:07
$begingroup$
+1. Note, however, that $sqrt{x^2}=|x|$, so you really should say a word or two about why you can write simply (and correctly) $sqrt{p^2(1-ecos t)^2} = p(1-ecos t)$. (Oh, and be sure to "accept" your answer.)
$endgroup$
– Blue
Jan 10 at 4:07
add a comment |
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Solve your definition of $e$ for $q^2$, and substitute that into the expressions for the distances.
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– Blue
Jan 10 at 1:39
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Thanks, @Blue! It took me awhile to see the nice factorization, but I finally did. Thanks again!
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– livingtolearn-learningtolive
Jan 10 at 3:23
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You should post your solution as an answer so that we can upvote it (and so that the question gets removed from the Unanswered queue).
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– Blue
Jan 10 at 3:25
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Done @blue. Thanks for you help again.
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– livingtolearn-learningtolive
Jan 10 at 4:01