How to show that the angle at intersections of concentric ellipses with a radial line is constant?












1












$begingroup$


I'm looking for a proof that the angle at intersections of concentric ellipses with a radial line remains constant as you move from one concentric ellipse to another.



For example, a series of concentric ellipses of different sizes are centered on a point of origin.
The radial lines from the point of origin outwards that trace the semi-major and semi-minor axis of the concentric ellipses only form right-angles at each intersection with the concentric ellipses.



How may one show that the other radial lines forming non-perpendicular intersections with the concentric circles form a constant angle with respect to the same radial line?






geometry conic-sections







share|cite|improve this question











$endgroup$












  • $begingroup$
    I hope my edit above clarifies and adds some context to the geometry proof problem. Thanks
    $endgroup$
    – Owen
    Dec 27 '18 at 10:15










  • $begingroup$
    To be clear: The angles are formed by the radial line and the tangent lines to the ellipses at the points of intersection, yes? If so, then "concentric" is clearly not enough of a constraint. A circle is an ellipse, and the tangent at the intersection with any radial line makes a right angle, where this isn't true for a non-circle ellipse. So, the ellipses need the same eccentricity. Moreover, you not only want the centers aligned, but you want the major and minor axes to be aligned; otherwise, one ellipse could be rotated with respect to another, throwing-off the property you describe.
    $endgroup$
    – Blue
    Dec 27 '18 at 10:23










  • $begingroup$
    I see, yes. The angles are between the radial line and the tangent lines to the ellipses at the points of intersection. The concentric ellipses also have the same eccentricity, and have semi-major and semi-minor axes aligned.
    $endgroup$
    – Owen
    Dec 27 '18 at 10:27












  • $begingroup$
    But then ... Ellipses with the same eccentricity are similar. Arranging them with a common center, and aligning their axes, we have that a simple dilation transformation carries any one ellipse to any other (and fixes any radial line); since dilations preserve angles, we're done.
    $endgroup$
    – Blue
    Dec 27 '18 at 10:27








  • 1




    $begingroup$
    The facts that all ellipses have the same eccentricity and that their axes are aligned are vital and should be edited into the question itself, not buried in the comments.
    $endgroup$
    – David K
    Dec 27 '18 at 12:56


















1












$begingroup$


I'm looking for a proof that the angle at intersections of concentric ellipses with a radial line remains constant as you move from one concentric ellipse to another.



For example, a series of concentric ellipses of different sizes are centered on a point of origin.
The radial lines from the point of origin outwards that trace the semi-major and semi-minor axis of the concentric ellipses only form right-angles at each intersection with the concentric ellipses.



How may one show that the other radial lines forming non-perpendicular intersections with the concentric circles form a constant angle with respect to the same radial line?






geometry conic-sections







share|cite|improve this question











$endgroup$












  • $begingroup$
    I hope my edit above clarifies and adds some context to the geometry proof problem. Thanks
    $endgroup$
    – Owen
    Dec 27 '18 at 10:15










  • $begingroup$
    To be clear: The angles are formed by the radial line and the tangent lines to the ellipses at the points of intersection, yes? If so, then "concentric" is clearly not enough of a constraint. A circle is an ellipse, and the tangent at the intersection with any radial line makes a right angle, where this isn't true for a non-circle ellipse. So, the ellipses need the same eccentricity. Moreover, you not only want the centers aligned, but you want the major and minor axes to be aligned; otherwise, one ellipse could be rotated with respect to another, throwing-off the property you describe.
    $endgroup$
    – Blue
    Dec 27 '18 at 10:23










  • $begingroup$
    I see, yes. The angles are between the radial line and the tangent lines to the ellipses at the points of intersection. The concentric ellipses also have the same eccentricity, and have semi-major and semi-minor axes aligned.
    $endgroup$
    – Owen
    Dec 27 '18 at 10:27












  • $begingroup$
    But then ... Ellipses with the same eccentricity are similar. Arranging them with a common center, and aligning their axes, we have that a simple dilation transformation carries any one ellipse to any other (and fixes any radial line); since dilations preserve angles, we're done.
    $endgroup$
    – Blue
    Dec 27 '18 at 10:27








  • 1




    $begingroup$
    The facts that all ellipses have the same eccentricity and that their axes are aligned are vital and should be edited into the question itself, not buried in the comments.
    $endgroup$
    – David K
    Dec 27 '18 at 12:56
















1












1








1





$begingroup$


I'm looking for a proof that the angle at intersections of concentric ellipses with a radial line remains constant as you move from one concentric ellipse to another.



For example, a series of concentric ellipses of different sizes are centered on a point of origin.
The radial lines from the point of origin outwards that trace the semi-major and semi-minor axis of the concentric ellipses only form right-angles at each intersection with the concentric ellipses.



How may one show that the other radial lines forming non-perpendicular intersections with the concentric circles form a constant angle with respect to the same radial line?






geometry conic-sections







share|cite|improve this question











$endgroup$




I'm looking for a proof that the angle at intersections of concentric ellipses with a radial line remains constant as you move from one concentric ellipse to another.



For example, a series of concentric ellipses of different sizes are centered on a point of origin.
The radial lines from the point of origin outwards that trace the semi-major and semi-minor axis of the concentric ellipses only form right-angles at each intersection with the concentric ellipses.



How may one show that the other radial lines forming non-perpendicular intersections with the concentric circles form a constant angle with respect to the same radial line?






geometry conic-sections




geometry conic-sections






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 27 '18 at 10:26







Owen

















asked Dec 27 '18 at 9:55









OwenOwen

62




62












  • $begingroup$
    I hope my edit above clarifies and adds some context to the geometry proof problem. Thanks
    $endgroup$
    – Owen
    Dec 27 '18 at 10:15










  • $begingroup$
    To be clear: The angles are formed by the radial line and the tangent lines to the ellipses at the points of intersection, yes? If so, then "concentric" is clearly not enough of a constraint. A circle is an ellipse, and the tangent at the intersection with any radial line makes a right angle, where this isn't true for a non-circle ellipse. So, the ellipses need the same eccentricity. Moreover, you not only want the centers aligned, but you want the major and minor axes to be aligned; otherwise, one ellipse could be rotated with respect to another, throwing-off the property you describe.
    $endgroup$
    – Blue
    Dec 27 '18 at 10:23










  • $begingroup$
    I see, yes. The angles are between the radial line and the tangent lines to the ellipses at the points of intersection. The concentric ellipses also have the same eccentricity, and have semi-major and semi-minor axes aligned.
    $endgroup$
    – Owen
    Dec 27 '18 at 10:27












  • $begingroup$
    But then ... Ellipses with the same eccentricity are similar. Arranging them with a common center, and aligning their axes, we have that a simple dilation transformation carries any one ellipse to any other (and fixes any radial line); since dilations preserve angles, we're done.
    $endgroup$
    – Blue
    Dec 27 '18 at 10:27








  • 1




    $begingroup$
    The facts that all ellipses have the same eccentricity and that their axes are aligned are vital and should be edited into the question itself, not buried in the comments.
    $endgroup$
    – David K
    Dec 27 '18 at 12:56




















  • $begingroup$
    I hope my edit above clarifies and adds some context to the geometry proof problem. Thanks
    $endgroup$
    – Owen
    Dec 27 '18 at 10:15










  • $begingroup$
    To be clear: The angles are formed by the radial line and the tangent lines to the ellipses at the points of intersection, yes? If so, then "concentric" is clearly not enough of a constraint. A circle is an ellipse, and the tangent at the intersection with any radial line makes a right angle, where this isn't true for a non-circle ellipse. So, the ellipses need the same eccentricity. Moreover, you not only want the centers aligned, but you want the major and minor axes to be aligned; otherwise, one ellipse could be rotated with respect to another, throwing-off the property you describe.
    $endgroup$
    – Blue
    Dec 27 '18 at 10:23










  • $begingroup$
    I see, yes. The angles are between the radial line and the tangent lines to the ellipses at the points of intersection. The concentric ellipses also have the same eccentricity, and have semi-major and semi-minor axes aligned.
    $endgroup$
    – Owen
    Dec 27 '18 at 10:27












  • $begingroup$
    But then ... Ellipses with the same eccentricity are similar. Arranging them with a common center, and aligning their axes, we have that a simple dilation transformation carries any one ellipse to any other (and fixes any radial line); since dilations preserve angles, we're done.
    $endgroup$
    – Blue
    Dec 27 '18 at 10:27








  • 1




    $begingroup$
    The facts that all ellipses have the same eccentricity and that their axes are aligned are vital and should be edited into the question itself, not buried in the comments.
    $endgroup$
    – David K
    Dec 27 '18 at 12:56


















$begingroup$
I hope my edit above clarifies and adds some context to the geometry proof problem. Thanks
$endgroup$
– Owen
Dec 27 '18 at 10:15




$begingroup$
I hope my edit above clarifies and adds some context to the geometry proof problem. Thanks
$endgroup$
– Owen
Dec 27 '18 at 10:15












$begingroup$
To be clear: The angles are formed by the radial line and the tangent lines to the ellipses at the points of intersection, yes? If so, then "concentric" is clearly not enough of a constraint. A circle is an ellipse, and the tangent at the intersection with any radial line makes a right angle, where this isn't true for a non-circle ellipse. So, the ellipses need the same eccentricity. Moreover, you not only want the centers aligned, but you want the major and minor axes to be aligned; otherwise, one ellipse could be rotated with respect to another, throwing-off the property you describe.
$endgroup$
– Blue
Dec 27 '18 at 10:23




$begingroup$
To be clear: The angles are formed by the radial line and the tangent lines to the ellipses at the points of intersection, yes? If so, then "concentric" is clearly not enough of a constraint. A circle is an ellipse, and the tangent at the intersection with any radial line makes a right angle, where this isn't true for a non-circle ellipse. So, the ellipses need the same eccentricity. Moreover, you not only want the centers aligned, but you want the major and minor axes to be aligned; otherwise, one ellipse could be rotated with respect to another, throwing-off the property you describe.
$endgroup$
– Blue
Dec 27 '18 at 10:23












$begingroup$
I see, yes. The angles are between the radial line and the tangent lines to the ellipses at the points of intersection. The concentric ellipses also have the same eccentricity, and have semi-major and semi-minor axes aligned.
$endgroup$
– Owen
Dec 27 '18 at 10:27






$begingroup$
I see, yes. The angles are between the radial line and the tangent lines to the ellipses at the points of intersection. The concentric ellipses also have the same eccentricity, and have semi-major and semi-minor axes aligned.
$endgroup$
– Owen
Dec 27 '18 at 10:27














$begingroup$
But then ... Ellipses with the same eccentricity are similar. Arranging them with a common center, and aligning their axes, we have that a simple dilation transformation carries any one ellipse to any other (and fixes any radial line); since dilations preserve angles, we're done.
$endgroup$
– Blue
Dec 27 '18 at 10:27






$begingroup$
But then ... Ellipses with the same eccentricity are similar. Arranging them with a common center, and aligning their axes, we have that a simple dilation transformation carries any one ellipse to any other (and fixes any radial line); since dilations preserve angles, we're done.
$endgroup$
– Blue
Dec 27 '18 at 10:27






1




1




$begingroup$
The facts that all ellipses have the same eccentricity and that their axes are aligned are vital and should be edited into the question itself, not buried in the comments.
$endgroup$
– David K
Dec 27 '18 at 12:56






$begingroup$
The facts that all ellipses have the same eccentricity and that their axes are aligned are vital and should be edited into the question itself, not buried in the comments.
$endgroup$
– David K
Dec 27 '18 at 12:56












1 Answer
1






active

oldest

votes


















0












$begingroup$

If someone is not convinced by dilation argument, you can also prove the fact in the following way:



The equation of ellipse is:



$$frac{x^2}{a^2}+frac{y^2}{b^2}=1$$



So the slope of the tangent is:



$$frac{x}{a^2}+frac{yy'}{b^2}=0$$



$$y'=-frac{b^2x}{a^2y}tag{1}$$



Now draw two ellipses with semi-axis $(a_1,b_1)$ and $(a_2,b_2)$. For concentric ellipses (with the same center and eccentricity):



$$frac{b_1}{a_1}=frac{b_2}{a_2}tag{2}$$



Draw a line $p$ passing through origin intersecting ellipses in points $(x_1,y_1)$ and $(x_2, y_2)$. Obviously:



$$frac{x_1}{y_1}=frac{x_2}{y_2}tag{3}$$



By combining (1), (2) and (3):



$$y_1'=-frac{b_1^2x_1}{a_1^2y_1}=-frac{b_2^2x_2}{a_2^2y_2}=y_2'$$



So both tangents have the same slopes which means that the tangents are parallel and therefore their angles with respect to line $p$ must be equal.



EDIT: I have assumed that concentric ellipses must have the same center and the same eccentricity. Eccentricity of an ellipse is defined with:



$$e=sqrt{1-frac{b^2}{a^2}}$$



So two ellipses with the same center are concentric only if (2) holds.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Equation $(2)$ asserts the similarity (and axis-alignment) of the ellipses, not the concentricity. Using the standard form of the ellipse equation guarantees that they have the same center.
    $endgroup$
    – Blue
    Dec 27 '18 at 12:01










  • $begingroup$
    @Blue: What is your definition of "concentric ellipses"?
    $endgroup$
    – Oldboy
    Dec 27 '18 at 12:16










  • $begingroup$
    "con-centric" = "same center".
    $endgroup$
    – Blue
    Dec 27 '18 at 12:24










  • $begingroup$
    True for circles, but that would mean that ellipses $x^2+y^2/100=1$ and $x^2/100+y^2=1$ are concentric too. It does not make any sense (at least to me) and it also invalidates the problem statement completely.
    $endgroup$
    – Oldboy
    Dec 27 '18 at 12:27








  • 1




    $begingroup$
    Whether the problem statement makes sense isn't relevant, as the author wasn't particularly precise in making that statement in the first place. Even so, I'd resist overloading "concentric" with shape information; we have "similar" to describe matching shapes. (Note that the problem also requires that the ellipses in question have the same orientation, which even the overloaded "concentric" misses ... unless we further overload the term to cover orientation.) Compare "confocal", used for arbitrarily-eccentric conics that share foci; "same shape"-ness is not part of the definition.
    $endgroup$
    – Blue
    Dec 27 '18 at 12:41













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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

If someone is not convinced by dilation argument, you can also prove the fact in the following way:



The equation of ellipse is:



$$frac{x^2}{a^2}+frac{y^2}{b^2}=1$$



So the slope of the tangent is:



$$frac{x}{a^2}+frac{yy'}{b^2}=0$$



$$y'=-frac{b^2x}{a^2y}tag{1}$$



Now draw two ellipses with semi-axis $(a_1,b_1)$ and $(a_2,b_2)$. For concentric ellipses (with the same center and eccentricity):



$$frac{b_1}{a_1}=frac{b_2}{a_2}tag{2}$$



Draw a line $p$ passing through origin intersecting ellipses in points $(x_1,y_1)$ and $(x_2, y_2)$. Obviously:



$$frac{x_1}{y_1}=frac{x_2}{y_2}tag{3}$$



By combining (1), (2) and (3):



$$y_1'=-frac{b_1^2x_1}{a_1^2y_1}=-frac{b_2^2x_2}{a_2^2y_2}=y_2'$$



So both tangents have the same slopes which means that the tangents are parallel and therefore their angles with respect to line $p$ must be equal.



EDIT: I have assumed that concentric ellipses must have the same center and the same eccentricity. Eccentricity of an ellipse is defined with:



$$e=sqrt{1-frac{b^2}{a^2}}$$



So two ellipses with the same center are concentric only if (2) holds.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Equation $(2)$ asserts the similarity (and axis-alignment) of the ellipses, not the concentricity. Using the standard form of the ellipse equation guarantees that they have the same center.
    $endgroup$
    – Blue
    Dec 27 '18 at 12:01










  • $begingroup$
    @Blue: What is your definition of "concentric ellipses"?
    $endgroup$
    – Oldboy
    Dec 27 '18 at 12:16










  • $begingroup$
    "con-centric" = "same center".
    $endgroup$
    – Blue
    Dec 27 '18 at 12:24










  • $begingroup$
    True for circles, but that would mean that ellipses $x^2+y^2/100=1$ and $x^2/100+y^2=1$ are concentric too. It does not make any sense (at least to me) and it also invalidates the problem statement completely.
    $endgroup$
    – Oldboy
    Dec 27 '18 at 12:27








  • 1




    $begingroup$
    Whether the problem statement makes sense isn't relevant, as the author wasn't particularly precise in making that statement in the first place. Even so, I'd resist overloading "concentric" with shape information; we have "similar" to describe matching shapes. (Note that the problem also requires that the ellipses in question have the same orientation, which even the overloaded "concentric" misses ... unless we further overload the term to cover orientation.) Compare "confocal", used for arbitrarily-eccentric conics that share foci; "same shape"-ness is not part of the definition.
    $endgroup$
    – Blue
    Dec 27 '18 at 12:41


















0












$begingroup$

If someone is not convinced by dilation argument, you can also prove the fact in the following way:



The equation of ellipse is:



$$frac{x^2}{a^2}+frac{y^2}{b^2}=1$$



So the slope of the tangent is:



$$frac{x}{a^2}+frac{yy'}{b^2}=0$$



$$y'=-frac{b^2x}{a^2y}tag{1}$$



Now draw two ellipses with semi-axis $(a_1,b_1)$ and $(a_2,b_2)$. For concentric ellipses (with the same center and eccentricity):



$$frac{b_1}{a_1}=frac{b_2}{a_2}tag{2}$$



Draw a line $p$ passing through origin intersecting ellipses in points $(x_1,y_1)$ and $(x_2, y_2)$. Obviously:



$$frac{x_1}{y_1}=frac{x_2}{y_2}tag{3}$$



By combining (1), (2) and (3):



$$y_1'=-frac{b_1^2x_1}{a_1^2y_1}=-frac{b_2^2x_2}{a_2^2y_2}=y_2'$$



So both tangents have the same slopes which means that the tangents are parallel and therefore their angles with respect to line $p$ must be equal.



EDIT: I have assumed that concentric ellipses must have the same center and the same eccentricity. Eccentricity of an ellipse is defined with:



$$e=sqrt{1-frac{b^2}{a^2}}$$



So two ellipses with the same center are concentric only if (2) holds.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Equation $(2)$ asserts the similarity (and axis-alignment) of the ellipses, not the concentricity. Using the standard form of the ellipse equation guarantees that they have the same center.
    $endgroup$
    – Blue
    Dec 27 '18 at 12:01










  • $begingroup$
    @Blue: What is your definition of "concentric ellipses"?
    $endgroup$
    – Oldboy
    Dec 27 '18 at 12:16










  • $begingroup$
    "con-centric" = "same center".
    $endgroup$
    – Blue
    Dec 27 '18 at 12:24










  • $begingroup$
    True for circles, but that would mean that ellipses $x^2+y^2/100=1$ and $x^2/100+y^2=1$ are concentric too. It does not make any sense (at least to me) and it also invalidates the problem statement completely.
    $endgroup$
    – Oldboy
    Dec 27 '18 at 12:27








  • 1




    $begingroup$
    Whether the problem statement makes sense isn't relevant, as the author wasn't particularly precise in making that statement in the first place. Even so, I'd resist overloading "concentric" with shape information; we have "similar" to describe matching shapes. (Note that the problem also requires that the ellipses in question have the same orientation, which even the overloaded "concentric" misses ... unless we further overload the term to cover orientation.) Compare "confocal", used for arbitrarily-eccentric conics that share foci; "same shape"-ness is not part of the definition.
    $endgroup$
    – Blue
    Dec 27 '18 at 12:41
















0












0








0





$begingroup$

If someone is not convinced by dilation argument, you can also prove the fact in the following way:



The equation of ellipse is:



$$frac{x^2}{a^2}+frac{y^2}{b^2}=1$$



So the slope of the tangent is:



$$frac{x}{a^2}+frac{yy'}{b^2}=0$$



$$y'=-frac{b^2x}{a^2y}tag{1}$$



Now draw two ellipses with semi-axis $(a_1,b_1)$ and $(a_2,b_2)$. For concentric ellipses (with the same center and eccentricity):



$$frac{b_1}{a_1}=frac{b_2}{a_2}tag{2}$$



Draw a line $p$ passing through origin intersecting ellipses in points $(x_1,y_1)$ and $(x_2, y_2)$. Obviously:



$$frac{x_1}{y_1}=frac{x_2}{y_2}tag{3}$$



By combining (1), (2) and (3):



$$y_1'=-frac{b_1^2x_1}{a_1^2y_1}=-frac{b_2^2x_2}{a_2^2y_2}=y_2'$$



So both tangents have the same slopes which means that the tangents are parallel and therefore their angles with respect to line $p$ must be equal.



EDIT: I have assumed that concentric ellipses must have the same center and the same eccentricity. Eccentricity of an ellipse is defined with:



$$e=sqrt{1-frac{b^2}{a^2}}$$



So two ellipses with the same center are concentric only if (2) holds.






share|cite|improve this answer











$endgroup$



If someone is not convinced by dilation argument, you can also prove the fact in the following way:



The equation of ellipse is:



$$frac{x^2}{a^2}+frac{y^2}{b^2}=1$$



So the slope of the tangent is:



$$frac{x}{a^2}+frac{yy'}{b^2}=0$$



$$y'=-frac{b^2x}{a^2y}tag{1}$$



Now draw two ellipses with semi-axis $(a_1,b_1)$ and $(a_2,b_2)$. For concentric ellipses (with the same center and eccentricity):



$$frac{b_1}{a_1}=frac{b_2}{a_2}tag{2}$$



Draw a line $p$ passing through origin intersecting ellipses in points $(x_1,y_1)$ and $(x_2, y_2)$. Obviously:



$$frac{x_1}{y_1}=frac{x_2}{y_2}tag{3}$$



By combining (1), (2) and (3):



$$y_1'=-frac{b_1^2x_1}{a_1^2y_1}=-frac{b_2^2x_2}{a_2^2y_2}=y_2'$$



So both tangents have the same slopes which means that the tangents are parallel and therefore their angles with respect to line $p$ must be equal.



EDIT: I have assumed that concentric ellipses must have the same center and the same eccentricity. Eccentricity of an ellipse is defined with:



$$e=sqrt{1-frac{b^2}{a^2}}$$



So two ellipses with the same center are concentric only if (2) holds.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 27 '18 at 12:21

























answered Dec 27 '18 at 11:51









OldboyOldboy

8,1651936




8,1651936












  • $begingroup$
    Equation $(2)$ asserts the similarity (and axis-alignment) of the ellipses, not the concentricity. Using the standard form of the ellipse equation guarantees that they have the same center.
    $endgroup$
    – Blue
    Dec 27 '18 at 12:01










  • $begingroup$
    @Blue: What is your definition of "concentric ellipses"?
    $endgroup$
    – Oldboy
    Dec 27 '18 at 12:16










  • $begingroup$
    "con-centric" = "same center".
    $endgroup$
    – Blue
    Dec 27 '18 at 12:24










  • $begingroup$
    True for circles, but that would mean that ellipses $x^2+y^2/100=1$ and $x^2/100+y^2=1$ are concentric too. It does not make any sense (at least to me) and it also invalidates the problem statement completely.
    $endgroup$
    – Oldboy
    Dec 27 '18 at 12:27








  • 1




    $begingroup$
    Whether the problem statement makes sense isn't relevant, as the author wasn't particularly precise in making that statement in the first place. Even so, I'd resist overloading "concentric" with shape information; we have "similar" to describe matching shapes. (Note that the problem also requires that the ellipses in question have the same orientation, which even the overloaded "concentric" misses ... unless we further overload the term to cover orientation.) Compare "confocal", used for arbitrarily-eccentric conics that share foci; "same shape"-ness is not part of the definition.
    $endgroup$
    – Blue
    Dec 27 '18 at 12:41




















  • $begingroup$
    Equation $(2)$ asserts the similarity (and axis-alignment) of the ellipses, not the concentricity. Using the standard form of the ellipse equation guarantees that they have the same center.
    $endgroup$
    – Blue
    Dec 27 '18 at 12:01










  • $begingroup$
    @Blue: What is your definition of "concentric ellipses"?
    $endgroup$
    – Oldboy
    Dec 27 '18 at 12:16










  • $begingroup$
    "con-centric" = "same center".
    $endgroup$
    – Blue
    Dec 27 '18 at 12:24










  • $begingroup$
    True for circles, but that would mean that ellipses $x^2+y^2/100=1$ and $x^2/100+y^2=1$ are concentric too. It does not make any sense (at least to me) and it also invalidates the problem statement completely.
    $endgroup$
    – Oldboy
    Dec 27 '18 at 12:27








  • 1




    $begingroup$
    Whether the problem statement makes sense isn't relevant, as the author wasn't particularly precise in making that statement in the first place. Even so, I'd resist overloading "concentric" with shape information; we have "similar" to describe matching shapes. (Note that the problem also requires that the ellipses in question have the same orientation, which even the overloaded "concentric" misses ... unless we further overload the term to cover orientation.) Compare "confocal", used for arbitrarily-eccentric conics that share foci; "same shape"-ness is not part of the definition.
    $endgroup$
    – Blue
    Dec 27 '18 at 12:41


















$begingroup$
Equation $(2)$ asserts the similarity (and axis-alignment) of the ellipses, not the concentricity. Using the standard form of the ellipse equation guarantees that they have the same center.
$endgroup$
– Blue
Dec 27 '18 at 12:01




$begingroup$
Equation $(2)$ asserts the similarity (and axis-alignment) of the ellipses, not the concentricity. Using the standard form of the ellipse equation guarantees that they have the same center.
$endgroup$
– Blue
Dec 27 '18 at 12:01












$begingroup$
@Blue: What is your definition of "concentric ellipses"?
$endgroup$
– Oldboy
Dec 27 '18 at 12:16




$begingroup$
@Blue: What is your definition of "concentric ellipses"?
$endgroup$
– Oldboy
Dec 27 '18 at 12:16












$begingroup$
"con-centric" = "same center".
$endgroup$
– Blue
Dec 27 '18 at 12:24




$begingroup$
"con-centric" = "same center".
$endgroup$
– Blue
Dec 27 '18 at 12:24












$begingroup$
True for circles, but that would mean that ellipses $x^2+y^2/100=1$ and $x^2/100+y^2=1$ are concentric too. It does not make any sense (at least to me) and it also invalidates the problem statement completely.
$endgroup$
– Oldboy
Dec 27 '18 at 12:27






$begingroup$
True for circles, but that would mean that ellipses $x^2+y^2/100=1$ and $x^2/100+y^2=1$ are concentric too. It does not make any sense (at least to me) and it also invalidates the problem statement completely.
$endgroup$
– Oldboy
Dec 27 '18 at 12:27






1




1




$begingroup$
Whether the problem statement makes sense isn't relevant, as the author wasn't particularly precise in making that statement in the first place. Even so, I'd resist overloading "concentric" with shape information; we have "similar" to describe matching shapes. (Note that the problem also requires that the ellipses in question have the same orientation, which even the overloaded "concentric" misses ... unless we further overload the term to cover orientation.) Compare "confocal", used for arbitrarily-eccentric conics that share foci; "same shape"-ness is not part of the definition.
$endgroup$
– Blue
Dec 27 '18 at 12:41






$begingroup$
Whether the problem statement makes sense isn't relevant, as the author wasn't particularly precise in making that statement in the first place. Even so, I'd resist overloading "concentric" with shape information; we have "similar" to describe matching shapes. (Note that the problem also requires that the ellipses in question have the same orientation, which even the overloaded "concentric" misses ... unless we further overload the term to cover orientation.) Compare "confocal", used for arbitrarily-eccentric conics that share foci; "same shape"-ness is not part of the definition.
$endgroup$
– Blue
Dec 27 '18 at 12:41




















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