Mixing
Prove that the semi Latus rectum of an ellipse is the harmonic mean of the segments of focal chord.
$begingroup$
I am a 12th student. I found this property in a reference book, without its proof.
So i tried to prove this myself but got stuck. Here's my attempt at the problem:
Basically the question is to prove $$frac{1}{AC} + frac{1}{AB} = frac{2a}{b^2}$$
Where $mathsf a$ and $mathsf b$ are semi-major and semi-minor axes, $mathsf CAB$ is the focal chord, $mathsf A $ is the focus and $mathsf AC$ and $mathsf AB$ are its segments.
I took parametric form for $mathsf B$ and $mathsf C$ as:
$$mathsf B (acosalpha, bsinalpha ) mathsf C (acosbeta, bsinbeta )$$
Where $alpha$ and $beta$ are eccentric angles of $mathsf B$ and $mathsf C$.
I found the length of segments of focal chord as:
$$AB = a(1 -ecosalpha )$$ and $$AC = a(1 -ecosbeta )$$ Where $A$ is $(ae, 0)$ and $e$ is the eccentricity.
After that i tried to apply harmonic mean but wasn't able to simplify the equation.
conic-sections coordinate-systems
asked Aug 4 '18 at 15:04
Pranav RajPranav Raj
63
$endgroup$
|
show 3 more comments
$begingroup$
I am a 12th student. I found this property in a reference book, without its proof.
So i tried to prove this myself but got stuck. Here's my attempt at the problem:
Basically the question is to prove $$frac{1}{AC} + frac{1}{AB} = frac{2a}{b^2}$$
Where $mathsf a$ and $mathsf b$ are semi-major and semi-minor axes, $mathsf CAB$ is the focal chord, $mathsf A $ is the focus and $mathsf AC$ and $mathsf AB$ are its segments.
I took parametric form for $mathsf B$ and $mathsf C$ as:
$$mathsf B (acosalpha, bsinalpha ) mathsf C (acosbeta, bsinbeta )$$
Where $alpha$ and $beta$ are eccentric angles of $mathsf B$ and $mathsf C$.
I found the length of segments of focal chord as:
$$AB = a(1 -ecosalpha )$$ and $$AC = a(1 -ecosbeta )$$ Where $A$ is $(ae, 0)$ and $e$ is the eccentricity.
After that i tried to apply harmonic mean but wasn't able to simplify the equation.
conic-sections coordinate-systems
asked Aug 4 '18 at 15:04
Pranav RajPranav Raj
63
$endgroup$
$begingroup$
Your Question would be greatly improved by including some details of your understanding of the problem. As it currently reads, it raises a suspicion that you do not make much sense of the terms used in the terse statement and are hoping someone will not only solve but explain for you what the terms mean.
$endgroup$
– hardmath
Aug 4 '18 at 17:14
$begingroup$
This is my first time asking a question. So forgive me for not being descriptive.I have edited my question accordingly.
$endgroup$
– Pranav Raj
Aug 5 '18 at 5:23
$begingroup$
how can i add a graph from geogebra to my post?
$endgroup$
– Pranav Raj
Aug 5 '18 at 5:27
1
$begingroup$
If you are familiar with polar equations, a solution is fairly immediate using the "polar form relative to a focus" equation. Wikipedia writes it in terms of the major axis and eccentricity, but it's a little cleaner (and more-relevant here) to use the semi-latus rectum, which I'll denote $ell$: $$r = frac{ell}{1+ecostheta}$$ (I've also somewhat arbitrarily chosen "$pm$" to be "$+$" in the denominator. This corresponds to putting the right-hand focus at the origin.)
$endgroup$
– Blue
Aug 5 '18 at 5:40
1
$begingroup$
See this answer, which addresses a separate question, but includes —in its "EDIT"— a nice proof of the harmonic mean property you seek. No coordinates (Cartesian or polar) needed, although the proof relies on the focus-directrix definition of a conic section.
$endgroup$
– Blue
Aug 5 '18 at 7:02
|
show 3 more comments
$begingroup$
I am a 12th student. I found this property in a reference book, without its proof.
So i tried to prove this myself but got stuck. Here's my attempt at the problem:
Basically the question is to prove $$frac{1}{AC} + frac{1}{AB} = frac{2a}{b^2}$$
Where $mathsf a$ and $mathsf b$ are semi-major and semi-minor axes, $mathsf CAB$ is the focal chord, $mathsf A $ is the focus and $mathsf AC$ and $mathsf AB$ are its segments.
I took parametric form for $mathsf B$ and $mathsf C$ as:
$$mathsf B (acosalpha, bsinalpha ) mathsf C (acosbeta, bsinbeta )$$
Where $alpha$ and $beta$ are eccentric angles of $mathsf B$ and $mathsf C$.
I found the length of segments of focal chord as:
$$AB = a(1 -ecosalpha )$$ and $$AC = a(1 -ecosbeta )$$ Where $A$ is $(ae, 0)$ and $e$ is the eccentricity.
After that i tried to apply harmonic mean but wasn't able to simplify the equation.
conic-sections coordinate-systems
asked Aug 4 '18 at 15:04
Pranav RajPranav Raj
63
$endgroup$
I am a 12th student. I found this property in a reference book, without its proof.
So i tried to prove this myself but got stuck. Here's my attempt at the problem:
Basically the question is to prove $$frac{1}{AC} + frac{1}{AB} = frac{2a}{b^2}$$
Where $mathsf a$ and $mathsf b$ are semi-major and semi-minor axes, $mathsf CAB$ is the focal chord, $mathsf A $ is the focus and $mathsf AC$ and $mathsf AB$ are its segments.
I took parametric form for $mathsf B$ and $mathsf C$ as:
$$mathsf B (acosalpha, bsinalpha ) mathsf C (acosbeta, bsinbeta )$$
Where $alpha$ and $beta$ are eccentric angles of $mathsf B$ and $mathsf C$.
I found the length of segments of focal chord as:
$$AB = a(1 -ecosalpha )$$ and $$AC = a(1 -ecosbeta )$$ Where $A$ is $(ae, 0)$ and $e$ is the eccentricity.
After that i tried to apply harmonic mean but wasn't able to simplify the equation.
conic-sections coordinate-systems
conic-sections coordinate-systems
asked Aug 4 '18 at 15:04
Pranav RajPranav Raj
63
asked Aug 4 '18 at 15:04
Pranav RajPranav Raj
63
edited Aug 5 '18 at 5:21
Pranav Raj
asked Aug 4 '18 at 15:04
Pranav RajPranav Raj
63
asked Aug 4 '18 at 15:04
Pranav RajPranav Raj
63
asked Aug 4 '18 at 15:04
Pranav RajPranav Raj
63
63
$begingroup$
Your Question would be greatly improved by including some details of your understanding of the problem. As it currently reads, it raises a suspicion that you do not make much sense of the terms used in the terse statement and are hoping someone will not only solve but explain for you what the terms mean.
$endgroup$
– hardmath
Aug 4 '18 at 17:14
$begingroup$
This is my first time asking a question. So forgive me for not being descriptive.I have edited my question accordingly.
$endgroup$
– Pranav Raj
Aug 5 '18 at 5:23
$begingroup$
how can i add a graph from geogebra to my post?
$endgroup$
– Pranav Raj
Aug 5 '18 at 5:27
1
$begingroup$
If you are familiar with polar equations, a solution is fairly immediate using the "polar form relative to a focus" equation. Wikipedia writes it in terms of the major axis and eccentricity, but it's a little cleaner (and more-relevant here) to use the semi-latus rectum, which I'll denote $ell$: $$r = frac{ell}{1+ecostheta}$$ (I've also somewhat arbitrarily chosen "$pm$" to be "$+$" in the denominator. This corresponds to putting the right-hand focus at the origin.)
$endgroup$
– Blue
Aug 5 '18 at 5:40
1
$begingroup$
See this answer, which addresses a separate question, but includes —in its "EDIT"— a nice proof of the harmonic mean property you seek. No coordinates (Cartesian or polar) needed, although the proof relies on the focus-directrix definition of a conic section.
$endgroup$
– Blue
Aug 5 '18 at 7:02
|
show 3 more comments
$begingroup$
Your Question would be greatly improved by including some details of your understanding of the problem. As it currently reads, it raises a suspicion that you do not make much sense of the terms used in the terse statement and are hoping someone will not only solve but explain for you what the terms mean.
$endgroup$
– hardmath
Aug 4 '18 at 17:14
$begingroup$
This is my first time asking a question. So forgive me for not being descriptive.I have edited my question accordingly.
$endgroup$
– Pranav Raj
Aug 5 '18 at 5:23
$begingroup$
how can i add a graph from geogebra to my post?
$endgroup$
– Pranav Raj
Aug 5 '18 at 5:27
1
$begingroup$
If you are familiar with polar equations, a solution is fairly immediate using the "polar form relative to a focus" equation. Wikipedia writes it in terms of the major axis and eccentricity, but it's a little cleaner (and more-relevant here) to use the semi-latus rectum, which I'll denote $ell$: $$r = frac{ell}{1+ecostheta}$$ (I've also somewhat arbitrarily chosen "$pm$" to be "$+$" in the denominator. This corresponds to putting the right-hand focus at the origin.)
$endgroup$
– Blue
Aug 5 '18 at 5:40
1
$begingroup$
See this answer, which addresses a separate question, but includes —in its "EDIT"— a nice proof of the harmonic mean property you seek. No coordinates (Cartesian or polar) needed, although the proof relies on the focus-directrix definition of a conic section.
$endgroup$
– Blue
Aug 5 '18 at 7:02
$begingroup$
Your Question would be greatly improved by including some details of your understanding of the problem. As it currently reads, it raises a suspicion that you do not make much sense of the terms used in the terse statement and are hoping someone will not only solve but explain for you what the terms mean.
$endgroup$
– hardmath
Aug 4 '18 at 17:14
$begingroup$
Your Question would be greatly improved by including some details of your understanding of the problem. As it currently reads, it raises a suspicion that you do not make much sense of the terms used in the terse statement and are hoping someone will not only solve but explain for you what the terms mean.
$endgroup$
– hardmath
Aug 4 '18 at 17:14
$begingroup$
This is my first time asking a question. So forgive me for not being descriptive.I have edited my question accordingly.
$endgroup$
– Pranav Raj
Aug 5 '18 at 5:23
$begingroup$
This is my first time asking a question. So forgive me for not being descriptive.I have edited my question accordingly.
$endgroup$
– Pranav Raj
Aug 5 '18 at 5:23
$begingroup$
how can i add a graph from geogebra to my post?
$endgroup$
– Pranav Raj
Aug 5 '18 at 5:27
$begingroup$
how can i add a graph from geogebra to my post?
$endgroup$
– Pranav Raj
Aug 5 '18 at 5:27
1
1
$begingroup$
If you are familiar with polar equations, a solution is fairly immediate using the "polar form relative to a focus" equation. Wikipedia writes it in terms of the major axis and eccentricity, but it's a little cleaner (and more-relevant here) to use the semi-latus rectum, which I'll denote $ell$: $$r = frac{ell}{1+ecostheta}$$ (I've also somewhat arbitrarily chosen "$pm$" to be "$+$" in the denominator. This corresponds to putting the right-hand focus at the origin.)
$endgroup$
– Blue
Aug 5 '18 at 5:40
$begingroup$
If you are familiar with polar equations, a solution is fairly immediate using the "polar form relative to a focus" equation. Wikipedia writes it in terms of the major axis and eccentricity, but it's a little cleaner (and more-relevant here) to use the semi-latus rectum, which I'll denote $ell$: $$r = frac{ell}{1+ecostheta}$$ (I've also somewhat arbitrarily chosen "$pm$" to be "$+$" in the denominator. This corresponds to putting the right-hand focus at the origin.)
$endgroup$
– Blue
Aug 5 '18 at 5:40
1
1
$begingroup$
See this answer, which addresses a separate question, but includes —in its "EDIT"— a nice proof of the harmonic mean property you seek. No coordinates (Cartesian or polar) needed, although the proof relies on the focus-directrix definition of a conic section.
$endgroup$
– Blue
Aug 5 '18 at 7:02
$begingroup$
See this answer, which addresses a separate question, but includes —in its "EDIT"— a nice proof of the harmonic mean property you seek. No coordinates (Cartesian or polar) needed, although the proof relies on the focus-directrix definition of a conic section.
$endgroup$
– Blue
Aug 5 '18 at 7:02
|
show 3 more comments
1 Answer
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$begingroup$
Let $O$ be a focus of a non-circle conic, let $overline{OR}$ be a semi-latus rectum, and let $overline{PQ}$ be a focal chord not containing $R$. Let $O^prime$, $P^prime$, $Q^prime$, $R^prime$ be the projections of $O$, $P$, $Q$, $R$, respectively, onto the directrix associated with $O$. Note that, by the focus-directrix property of a conic with eccentricity $eneq 0$,
$$e,|OX|=|XX^prime| quadtext{for}quad X=P, Q, R tag{1}$$
Defining $p:=|OP|$, $q:=|OQ|$, $r:=|OR|$, we have, by obvious similar triangles in the figure,
$$frac{(p-r)/e}{p} = frac{(r-q)/e}{q} qquadtoqquad r = frac{2pq}{p+q} = frac{2}{dfrac{1}{p}+dfrac{1}{q}} tag{$star$}$$
proving the result. $square$
answered Jan 1 at 16:09
BlueBlue
47.7k870151
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add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
votes
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oldest
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$begingroup$
Let $O$ be a focus of a non-circle conic, let $overline{OR}$ be a semi-latus rectum, and let $overline{PQ}$ be a focal chord not containing $R$. Let $O^prime$, $P^prime$, $Q^prime$, $R^prime$ be the projections of $O$, $P$, $Q$, $R$, respectively, onto the directrix associated with $O$. Note that, by the focus-directrix property of a conic with eccentricity $eneq 0$,
$$e,|OX|=|XX^prime| quadtext{for}quad X=P, Q, R tag{1}$$
Defining $p:=|OP|$, $q:=|OQ|$, $r:=|OR|$, we have, by obvious similar triangles in the figure,
$$frac{(p-r)/e}{p} = frac{(r-q)/e}{q} qquadtoqquad r = frac{2pq}{p+q} = frac{2}{dfrac{1}{p}+dfrac{1}{q}} tag{$star$}$$
proving the result. $square$
answered Jan 1 at 16:09
BlueBlue
47.7k870151
$endgroup$
add a comment |
$begingroup$
Let $O$ be a focus of a non-circle conic, let $overline{OR}$ be a semi-latus rectum, and let $overline{PQ}$ be a focal chord not containing $R$. Let $O^prime$, $P^prime$, $Q^prime$, $R^prime$ be the projections of $O$, $P$, $Q$, $R$, respectively, onto the directrix associated with $O$. Note that, by the focus-directrix property of a conic with eccentricity $eneq 0$,
$$e,|OX|=|XX^prime| quadtext{for}quad X=P, Q, R tag{1}$$
Defining $p:=|OP|$, $q:=|OQ|$, $r:=|OR|$, we have, by obvious similar triangles in the figure,
$$frac{(p-r)/e}{p} = frac{(r-q)/e}{q} qquadtoqquad r = frac{2pq}{p+q} = frac{2}{dfrac{1}{p}+dfrac{1}{q}} tag{$star$}$$
proving the result. $square$
answered Jan 1 at 16:09
BlueBlue
47.7k870151
$endgroup$
add a comment |
$begingroup$
Let $O$ be a focus of a non-circle conic, let $overline{OR}$ be a semi-latus rectum, and let $overline{PQ}$ be a focal chord not containing $R$. Let $O^prime$, $P^prime$, $Q^prime$, $R^prime$ be the projections of $O$, $P$, $Q$, $R$, respectively, onto the directrix associated with $O$. Note that, by the focus-directrix property of a conic with eccentricity $eneq 0$,
$$e,|OX|=|XX^prime| quadtext{for}quad X=P, Q, R tag{1}$$
Defining $p:=|OP|$, $q:=|OQ|$, $r:=|OR|$, we have, by obvious similar triangles in the figure,
$$frac{(p-r)/e}{p} = frac{(r-q)/e}{q} qquadtoqquad r = frac{2pq}{p+q} = frac{2}{dfrac{1}{p}+dfrac{1}{q}} tag{$star$}$$
proving the result. $square$
answered Jan 1 at 16:09
BlueBlue
47.7k870151
$endgroup$
Let $O$ be a focus of a non-circle conic, let $overline{OR}$ be a semi-latus rectum, and let $overline{PQ}$ be a focal chord not containing $R$. Let $O^prime$, $P^prime$, $Q^prime$, $R^prime$ be the projections of $O$, $P$, $Q$, $R$, respectively, onto the directrix associated with $O$. Note that, by the focus-directrix property of a conic with eccentricity $eneq 0$,
$$e,|OX|=|XX^prime| quadtext{for}quad X=P, Q, R tag{1}$$
Defining $p:=|OP|$, $q:=|OQ|$, $r:=|OR|$, we have, by obvious similar triangles in the figure,
$$frac{(p-r)/e}{p} = frac{(r-q)/e}{q} qquadtoqquad r = frac{2pq}{p+q} = frac{2}{dfrac{1}{p}+dfrac{1}{q}} tag{$star$}$$
proving the result. $square$
answered Jan 1 at 16:09
BlueBlue
47.7k870151
answered Jan 1 at 16:09
BlueBlue
47.7k870151
answered Jan 1 at 16:09
BlueBlue
47.7k870151
answered Jan 1 at 16:09
BlueBlue
47.7k870151
47.7k870151
add a comment |
add a comment |
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$begingroup$
Your Question would be greatly improved by including some details of your understanding of the problem. As it currently reads, it raises a suspicion that you do not make much sense of the terms used in the terse statement and are hoping someone will not only solve but explain for you what the terms mean.
$endgroup$
– hardmath
Aug 4 '18 at 17:14
$begingroup$
This is my first time asking a question. So forgive me for not being descriptive.I have edited my question accordingly.
$endgroup$
– Pranav Raj
Aug 5 '18 at 5:23
$begingroup$
how can i add a graph from geogebra to my post?
$endgroup$
– Pranav Raj
Aug 5 '18 at 5:27
1
$begingroup$
If you are familiar with polar equations, a solution is fairly immediate using the "polar form relative to a focus" equation. Wikipedia writes it in terms of the major axis and eccentricity, but it's a little cleaner (and more-relevant here) to use the semi-latus rectum, which I'll denote $ell$: $$r = frac{ell}{1+ecostheta}$$ (I've also somewhat arbitrarily chosen "$pm$" to be "$+$" in the denominator. This corresponds to putting the right-hand focus at the origin.)
$endgroup$
– Blue
Aug 5 '18 at 5:40
1
$begingroup$
See this answer, which addresses a separate question, but includes —in its "EDIT"— a nice proof of the harmonic mean property you seek. No coordinates (Cartesian or polar) needed, although the proof relies on the focus-directrix definition of a conic section.
$endgroup$
– Blue
Aug 5 '18 at 7:02