Why are two definitions of ellipses equivalent?
$begingroup$
In classical geometry an ellipse is usually defined as the locus of points in the plane such that the distances from each point to the two foci have a given sum.
When we speak of an ellipse analytically, we usually describe it as a circle that has been squashed in one direction, i.e. something similar to the curve $x^2+(y/b)^2 = 1$.
"Everyone knows" that these two definitions yield the same family of shapes. But how can that be proved?
geometry analytic-geometry conic-sections
asked Oct 29 '13 at 10:28
Henning MakholmHenning Makholm
241k17308547
$endgroup$
add a comment |
$begingroup$
In classical geometry an ellipse is usually defined as the locus of points in the plane such that the distances from each point to the two foci have a given sum.
When we speak of an ellipse analytically, we usually describe it as a circle that has been squashed in one direction, i.e. something similar to the curve $x^2+(y/b)^2 = 1$.
"Everyone knows" that these two definitions yield the same family of shapes. But how can that be proved?
geometry analytic-geometry conic-sections
asked Oct 29 '13 at 10:28
Henning MakholmHenning Makholm
241k17308547
$endgroup$
$begingroup$
I needed this fact when answering this question, but we don't seem to have a question specifically for this basic fact. My first attempt at proving it by mindless formula-crunching led me horribly astray, so it seems to be useful to have a way that works written down explicitly.
$endgroup$
– Henning Makholm
Oct 29 '13 at 10:29
$begingroup$
I'm interested in getting more answers to this, especially various intuitive explanations. Is it okay with you if I edit this question into a big-list question asking for more answers?
$endgroup$
– Tanner Swett
Apr 7 '17 at 3:05
1
$begingroup$
3Blue1Brown recently posted a video giving a beautiful geometric proof of the relationship between the 3 most common descriptions of ellipses.
$endgroup$
– Paul Sinclair
Jan 4 at 0:38
add a comment |
$begingroup$
In classical geometry an ellipse is usually defined as the locus of points in the plane such that the distances from each point to the two foci have a given sum.
When we speak of an ellipse analytically, we usually describe it as a circle that has been squashed in one direction, i.e. something similar to the curve $x^2+(y/b)^2 = 1$.
"Everyone knows" that these two definitions yield the same family of shapes. But how can that be proved?
geometry analytic-geometry conic-sections
asked Oct 29 '13 at 10:28
Henning MakholmHenning Makholm
241k17308547
$endgroup$
In classical geometry an ellipse is usually defined as the locus of points in the plane such that the distances from each point to the two foci have a given sum.
When we speak of an ellipse analytically, we usually describe it as a circle that has been squashed in one direction, i.e. something similar to the curve $x^2+(y/b)^2 = 1$.
"Everyone knows" that these two definitions yield the same family of shapes. But how can that be proved?
geometry analytic-geometry conic-sections
geometry analytic-geometry conic-sections
asked Oct 29 '13 at 10:28
Henning MakholmHenning Makholm
241k17308547
asked Oct 29 '13 at 10:28
Henning MakholmHenning Makholm
241k17308547
asked Oct 29 '13 at 10:28
Henning MakholmHenning Makholm
241k17308547
asked Oct 29 '13 at 10:28
Henning MakholmHenning Makholm
241k17308547
asked Oct 29 '13 at 10:28
Henning MakholmHenning Makholm
241k17308547
241k17308547
$begingroup$
I needed this fact when answering this question, but we don't seem to have a question specifically for this basic fact. My first attempt at proving it by mindless formula-crunching led me horribly astray, so it seems to be useful to have a way that works written down explicitly.
$endgroup$
– Henning Makholm
Oct 29 '13 at 10:29
$begingroup$
I'm interested in getting more answers to this, especially various intuitive explanations. Is it okay with you if I edit this question into a big-list question asking for more answers?
$endgroup$
– Tanner Swett
Apr 7 '17 at 3:05
1
$begingroup$
3Blue1Brown recently posted a video giving a beautiful geometric proof of the relationship between the 3 most common descriptions of ellipses.
$endgroup$
– Paul Sinclair
Jan 4 at 0:38
add a comment |
$begingroup$
I needed this fact when answering this question, but we don't seem to have a question specifically for this basic fact. My first attempt at proving it by mindless formula-crunching led me horribly astray, so it seems to be useful to have a way that works written down explicitly.
$endgroup$
– Henning Makholm
Oct 29 '13 at 10:29
$begingroup$
I'm interested in getting more answers to this, especially various intuitive explanations. Is it okay with you if I edit this question into a big-list question asking for more answers?
$endgroup$
– Tanner Swett
Apr 7 '17 at 3:05
1
$begingroup$
3Blue1Brown recently posted a video giving a beautiful geometric proof of the relationship between the 3 most common descriptions of ellipses.
$endgroup$
– Paul Sinclair
Jan 4 at 0:38
$begingroup$
I needed this fact when answering this question, but we don't seem to have a question specifically for this basic fact. My first attempt at proving it by mindless formula-crunching led me horribly astray, so it seems to be useful to have a way that works written down explicitly.
$endgroup$
– Henning Makholm
Oct 29 '13 at 10:29
$begingroup$
I needed this fact when answering this question, but we don't seem to have a question specifically for this basic fact. My first attempt at proving it by mindless formula-crunching led me horribly astray, so it seems to be useful to have a way that works written down explicitly.
$endgroup$
– Henning Makholm
Oct 29 '13 at 10:29
$begingroup$
I'm interested in getting more answers to this, especially various intuitive explanations. Is it okay with you if I edit this question into a big-list question asking for more answers?
$endgroup$
– Tanner Swett
Apr 7 '17 at 3:05
$begingroup$
I'm interested in getting more answers to this, especially various intuitive explanations. Is it okay with you if I edit this question into a big-list question asking for more answers?
$endgroup$
– Tanner Swett
Apr 7 '17 at 3:05
1
1
$begingroup$
3Blue1Brown recently posted a video giving a beautiful geometric proof of the relationship between the 3 most common descriptions of ellipses.
$endgroup$
– Paul Sinclair
Jan 4 at 0:38
$begingroup$
3Blue1Brown recently posted a video giving a beautiful geometric proof of the relationship between the 3 most common descriptions of ellipses.
$endgroup$
– Paul Sinclair
Jan 4 at 0:38
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Suppose we have a classical ellipse with its two foci and major axis given. We want to prove that it's a squashed circle.
Select a coordinate system with its origin in the center if the ellipse, the $x$-axis passing through the foci, and a scale such that the common sum-of-distances-to-the-foci is $2$. The foci have coordinates $(pm c,0)$ for some $cin[0,1)$.
By considering the nodes at the end of the minor axis we find that the semiminor axis of the ellipse must be $b=sqrt{1-c^2}$. We must then prove that the equations
$$ tag{1} x^2 + left(frac{y}{sqrt{1-c^2}}right)^2 = 1 $$
$$ tag{2} sqrt{(x+c)^2+y^2} + sqrt{(x-c)^2+y^2} = 2 $$
are equivalent.
Rearranging (1) gives
$$tag{1'} y^2 = (1-c^2)(1-x^2) $$
and therefore $(x+c)^2+y^2 = (1+xc)^2$ by multiplying out each side and collecting terms. Similarly, $(x-c)^2+y^2=(1-xc)^2$. So for points where (1) holds, (2) reduces to $(1+xc)+(1-xc)=2$, which is of course true.
Thus every solution of (1) is a solution of (2). On the other hand, for every fixed $xin(-1,1)$, it is clear that the left-hand-side of (2) increases monotonically with $|y|$ so it can have at most two solutions, which must then be exactly the two solutions for $y$ we get from (1'). (And neither equation has solutions with $|x|>1$). So the equations are indeed equivalent.
Conversely, if we have a squashed circle with major and minor axes $2a$ and $2b$, we can find the focal distance $2c$ by $c^2+b^2=a^2$ and locate the foci on the major axis. Then the above argument shows that the classical ellipse with these foci coincides with our squashed circle.
answered Oct 29 '13 at 10:28
Henning MakholmHenning Makholm
241k17308547
$endgroup$
$begingroup$
Awesome algebraic explanation! However, do you think there's a way to better see this connection geometrically? To somehow see that if we stretch a circle, there will be two points whose sum of the distances to any point on the ellipse will be a constant? I've been trying to come up with one and haven't been able to. Thank you!
$endgroup$
– Joshua Ronis
Jan 5 at 18:51
$begingroup$
@JoshuaRonis: If I had a good geometric proof, I wouldn't have needed to rely on mindless formula-crunching here.
$endgroup$
– Henning Makholm
Jan 6 at 14:38
$begingroup$
Haha okay I'm asking in a bunch of forums and trying to find one myself, I'll post here if I do
$endgroup$
– Joshua Ronis
Jan 6 at 15:30
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
Suppose we have a classical ellipse with its two foci and major axis given. We want to prove that it's a squashed circle.
Select a coordinate system with its origin in the center if the ellipse, the $x$-axis passing through the foci, and a scale such that the common sum-of-distances-to-the-foci is $2$. The foci have coordinates $(pm c,0)$ for some $cin[0,1)$.
By considering the nodes at the end of the minor axis we find that the semiminor axis of the ellipse must be $b=sqrt{1-c^2}$. We must then prove that the equations
$$ tag{1} x^2 + left(frac{y}{sqrt{1-c^2}}right)^2 = 1 $$
$$ tag{2} sqrt{(x+c)^2+y^2} + sqrt{(x-c)^2+y^2} = 2 $$
are equivalent.
Rearranging (1) gives
$$tag{1'} y^2 = (1-c^2)(1-x^2) $$
and therefore $(x+c)^2+y^2 = (1+xc)^2$ by multiplying out each side and collecting terms. Similarly, $(x-c)^2+y^2=(1-xc)^2$. So for points where (1) holds, (2) reduces to $(1+xc)+(1-xc)=2$, which is of course true.
Thus every solution of (1) is a solution of (2). On the other hand, for every fixed $xin(-1,1)$, it is clear that the left-hand-side of (2) increases monotonically with $|y|$ so it can have at most two solutions, which must then be exactly the two solutions for $y$ we get from (1'). (And neither equation has solutions with $|x|>1$). So the equations are indeed equivalent.
Conversely, if we have a squashed circle with major and minor axes $2a$ and $2b$, we can find the focal distance $2c$ by $c^2+b^2=a^2$ and locate the foci on the major axis. Then the above argument shows that the classical ellipse with these foci coincides with our squashed circle.
answered Oct 29 '13 at 10:28
Henning MakholmHenning Makholm
241k17308547
$endgroup$
$begingroup$
Awesome algebraic explanation! However, do you think there's a way to better see this connection geometrically? To somehow see that if we stretch a circle, there will be two points whose sum of the distances to any point on the ellipse will be a constant? I've been trying to come up with one and haven't been able to. Thank you!
$endgroup$
– Joshua Ronis
Jan 5 at 18:51
$begingroup$
@JoshuaRonis: If I had a good geometric proof, I wouldn't have needed to rely on mindless formula-crunching here.
$endgroup$
– Henning Makholm
Jan 6 at 14:38
$begingroup$
Haha okay I'm asking in a bunch of forums and trying to find one myself, I'll post here if I do
$endgroup$
– Joshua Ronis
Jan 6 at 15:30
add a comment |
$begingroup$
Suppose we have a classical ellipse with its two foci and major axis given. We want to prove that it's a squashed circle.
Select a coordinate system with its origin in the center if the ellipse, the $x$-axis passing through the foci, and a scale such that the common sum-of-distances-to-the-foci is $2$. The foci have coordinates $(pm c,0)$ for some $cin[0,1)$.
By considering the nodes at the end of the minor axis we find that the semiminor axis of the ellipse must be $b=sqrt{1-c^2}$. We must then prove that the equations
$$ tag{1} x^2 + left(frac{y}{sqrt{1-c^2}}right)^2 = 1 $$
$$ tag{2} sqrt{(x+c)^2+y^2} + sqrt{(x-c)^2+y^2} = 2 $$
are equivalent.
Rearranging (1) gives
$$tag{1'} y^2 = (1-c^2)(1-x^2) $$
and therefore $(x+c)^2+y^2 = (1+xc)^2$ by multiplying out each side and collecting terms. Similarly, $(x-c)^2+y^2=(1-xc)^2$. So for points where (1) holds, (2) reduces to $(1+xc)+(1-xc)=2$, which is of course true.
Thus every solution of (1) is a solution of (2). On the other hand, for every fixed $xin(-1,1)$, it is clear that the left-hand-side of (2) increases monotonically with $|y|$ so it can have at most two solutions, which must then be exactly the two solutions for $y$ we get from (1'). (And neither equation has solutions with $|x|>1$). So the equations are indeed equivalent.
Conversely, if we have a squashed circle with major and minor axes $2a$ and $2b$, we can find the focal distance $2c$ by $c^2+b^2=a^2$ and locate the foci on the major axis. Then the above argument shows that the classical ellipse with these foci coincides with our squashed circle.
answered Oct 29 '13 at 10:28
Henning MakholmHenning Makholm
241k17308547
$endgroup$
$begingroup$
Awesome algebraic explanation! However, do you think there's a way to better see this connection geometrically? To somehow see that if we stretch a circle, there will be two points whose sum of the distances to any point on the ellipse will be a constant? I've been trying to come up with one and haven't been able to. Thank you!
$endgroup$
– Joshua Ronis
Jan 5 at 18:51
$begingroup$
@JoshuaRonis: If I had a good geometric proof, I wouldn't have needed to rely on mindless formula-crunching here.
$endgroup$
– Henning Makholm
Jan 6 at 14:38
$begingroup$
Haha okay I'm asking in a bunch of forums and trying to find one myself, I'll post here if I do
$endgroup$
– Joshua Ronis
Jan 6 at 15:30
add a comment |
$begingroup$
Suppose we have a classical ellipse with its two foci and major axis given. We want to prove that it's a squashed circle.
Select a coordinate system with its origin in the center if the ellipse, the $x$-axis passing through the foci, and a scale such that the common sum-of-distances-to-the-foci is $2$. The foci have coordinates $(pm c,0)$ for some $cin[0,1)$.
By considering the nodes at the end of the minor axis we find that the semiminor axis of the ellipse must be $b=sqrt{1-c^2}$. We must then prove that the equations
$$ tag{1} x^2 + left(frac{y}{sqrt{1-c^2}}right)^2 = 1 $$
$$ tag{2} sqrt{(x+c)^2+y^2} + sqrt{(x-c)^2+y^2} = 2 $$
are equivalent.
Rearranging (1) gives
$$tag{1'} y^2 = (1-c^2)(1-x^2) $$
and therefore $(x+c)^2+y^2 = (1+xc)^2$ by multiplying out each side and collecting terms. Similarly, $(x-c)^2+y^2=(1-xc)^2$. So for points where (1) holds, (2) reduces to $(1+xc)+(1-xc)=2$, which is of course true.
Thus every solution of (1) is a solution of (2). On the other hand, for every fixed $xin(-1,1)$, it is clear that the left-hand-side of (2) increases monotonically with $|y|$ so it can have at most two solutions, which must then be exactly the two solutions for $y$ we get from (1'). (And neither equation has solutions with $|x|>1$). So the equations are indeed equivalent.
Conversely, if we have a squashed circle with major and minor axes $2a$ and $2b$, we can find the focal distance $2c$ by $c^2+b^2=a^2$ and locate the foci on the major axis. Then the above argument shows that the classical ellipse with these foci coincides with our squashed circle.
answered Oct 29 '13 at 10:28
Henning MakholmHenning Makholm
241k17308547
$endgroup$
Suppose we have a classical ellipse with its two foci and major axis given. We want to prove that it's a squashed circle.
Select a coordinate system with its origin in the center if the ellipse, the $x$-axis passing through the foci, and a scale such that the common sum-of-distances-to-the-foci is $2$. The foci have coordinates $(pm c,0)$ for some $cin[0,1)$.
By considering the nodes at the end of the minor axis we find that the semiminor axis of the ellipse must be $b=sqrt{1-c^2}$. We must then prove that the equations
$$ tag{1} x^2 + left(frac{y}{sqrt{1-c^2}}right)^2 = 1 $$
$$ tag{2} sqrt{(x+c)^2+y^2} + sqrt{(x-c)^2+y^2} = 2 $$
are equivalent.
Rearranging (1) gives
$$tag{1'} y^2 = (1-c^2)(1-x^2) $$
and therefore $(x+c)^2+y^2 = (1+xc)^2$ by multiplying out each side and collecting terms. Similarly, $(x-c)^2+y^2=(1-xc)^2$. So for points where (1) holds, (2) reduces to $(1+xc)+(1-xc)=2$, which is of course true.
Thus every solution of (1) is a solution of (2). On the other hand, for every fixed $xin(-1,1)$, it is clear that the left-hand-side of (2) increases monotonically with $|y|$ so it can have at most two solutions, which must then be exactly the two solutions for $y$ we get from (1'). (And neither equation has solutions with $|x|>1$). So the equations are indeed equivalent.
Conversely, if we have a squashed circle with major and minor axes $2a$ and $2b$, we can find the focal distance $2c$ by $c^2+b^2=a^2$ and locate the foci on the major axis. Then the above argument shows that the classical ellipse with these foci coincides with our squashed circle.
answered Oct 29 '13 at 10:28
Henning MakholmHenning Makholm
241k17308547
answered Oct 29 '13 at 10:28
Henning MakholmHenning Makholm
241k17308547
answered Oct 29 '13 at 10:28
Henning MakholmHenning Makholm
241k17308547
answered Oct 29 '13 at 10:28
Henning MakholmHenning Makholm
241k17308547
241k17308547
$begingroup$
Awesome algebraic explanation! However, do you think there's a way to better see this connection geometrically? To somehow see that if we stretch a circle, there will be two points whose sum of the distances to any point on the ellipse will be a constant? I've been trying to come up with one and haven't been able to. Thank you!
$endgroup$
– Joshua Ronis
Jan 5 at 18:51
$begingroup$
@JoshuaRonis: If I had a good geometric proof, I wouldn't have needed to rely on mindless formula-crunching here.
$endgroup$
– Henning Makholm
Jan 6 at 14:38
$begingroup$
Haha okay I'm asking in a bunch of forums and trying to find one myself, I'll post here if I do
$endgroup$
– Joshua Ronis
Jan 6 at 15:30
add a comment |
$begingroup$
Awesome algebraic explanation! However, do you think there's a way to better see this connection geometrically? To somehow see that if we stretch a circle, there will be two points whose sum of the distances to any point on the ellipse will be a constant? I've been trying to come up with one and haven't been able to. Thank you!
$endgroup$
– Joshua Ronis
Jan 5 at 18:51
$begingroup$
@JoshuaRonis: If I had a good geometric proof, I wouldn't have needed to rely on mindless formula-crunching here.
$endgroup$
– Henning Makholm
Jan 6 at 14:38
$begingroup$
Haha okay I'm asking in a bunch of forums and trying to find one myself, I'll post here if I do
$endgroup$
– Joshua Ronis
Jan 6 at 15:30
$begingroup$
Awesome algebraic explanation! However, do you think there's a way to better see this connection geometrically? To somehow see that if we stretch a circle, there will be two points whose sum of the distances to any point on the ellipse will be a constant? I've been trying to come up with one and haven't been able to. Thank you!
$endgroup$
– Joshua Ronis
Jan 5 at 18:51
$begingroup$
Awesome algebraic explanation! However, do you think there's a way to better see this connection geometrically? To somehow see that if we stretch a circle, there will be two points whose sum of the distances to any point on the ellipse will be a constant? I've been trying to come up with one and haven't been able to. Thank you!
$endgroup$
– Joshua Ronis
Jan 5 at 18:51
$begingroup$
@JoshuaRonis: If I had a good geometric proof, I wouldn't have needed to rely on mindless formula-crunching here.
$endgroup$
– Henning Makholm
Jan 6 at 14:38
$begingroup$
@JoshuaRonis: If I had a good geometric proof, I wouldn't have needed to rely on mindless formula-crunching here.
$endgroup$
– Henning Makholm
Jan 6 at 14:38
$begingroup$
Haha okay I'm asking in a bunch of forums and trying to find one myself, I'll post here if I do
$endgroup$
– Joshua Ronis
Jan 6 at 15:30
$begingroup$
Haha okay I'm asking in a bunch of forums and trying to find one myself, I'll post here if I do
$endgroup$
– Joshua Ronis
Jan 6 at 15:30
add a comment |
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$begingroup$
I needed this fact when answering this question, but we don't seem to have a question specifically for this basic fact. My first attempt at proving it by mindless formula-crunching led me horribly astray, so it seems to be useful to have a way that works written down explicitly.
$endgroup$
– Henning Makholm
Oct 29 '13 at 10:29
$begingroup$
I'm interested in getting more answers to this, especially various intuitive explanations. Is it okay with you if I edit this question into a big-list question asking for more answers?
$endgroup$
– Tanner Swett
Apr 7 '17 at 3:05
1
$begingroup$
3Blue1Brown recently posted a video giving a beautiful geometric proof of the relationship between the 3 most common descriptions of ellipses.
$endgroup$
– Paul Sinclair
Jan 4 at 0:38