Does the curve $2^2 x^2 + 4^2 y^2 = (x^2 + y^2)^2$ pass through $(0,0)$?
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Any ideas about why Desmos/Wolfram Alpha) does not show $(0,0)$ as a part of the curve $$2^2 x^2 + 4^2 y^2 = (x^2 + y^2)^2$$ will be appreciated.
Am I missing something or is it a bug?
graphing-functions curves
edited Dec 15 '18 at 12:52
amWhy
1
asked Dec 15 '18 at 12:16
Winged Blades of GodricWinged Blades of Godric
665
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add a comment |
$begingroup$
Any ideas about why Desmos/Wolfram Alpha) does not show $(0,0)$ as a part of the curve $$2^2 x^2 + 4^2 y^2 = (x^2 + y^2)^2$$ will be appreciated.
Am I missing something or is it a bug?
graphing-functions curves
edited Dec 15 '18 at 12:52
amWhy
1
asked Dec 15 '18 at 12:16
Winged Blades of GodricWinged Blades of Godric
665
$endgroup$
add a comment |
$begingroup$
Any ideas about why Desmos/Wolfram Alpha) does not show $(0,0)$ as a part of the curve $$2^2 x^2 + 4^2 y^2 = (x^2 + y^2)^2$$ will be appreciated.
Am I missing something or is it a bug?
graphing-functions curves
edited Dec 15 '18 at 12:52
amWhy
1
asked Dec 15 '18 at 12:16
Winged Blades of GodricWinged Blades of Godric
665
$endgroup$
Any ideas about why Desmos/Wolfram Alpha) does not show $(0,0)$ as a part of the curve $$2^2 x^2 + 4^2 y^2 = (x^2 + y^2)^2$$ will be appreciated.
Am I missing something or is it a bug?
graphing-functions curves
graphing-functions curves
edited Dec 15 '18 at 12:52
amWhy
1
asked Dec 15 '18 at 12:16
Winged Blades of GodricWinged Blades of Godric
665
edited Dec 15 '18 at 12:52
amWhy
1
asked Dec 15 '18 at 12:16
Winged Blades of GodricWinged Blades of Godric
665
edited Dec 15 '18 at 12:52
amWhy
1
edited Dec 15 '18 at 12:52
amWhy
1
edited Dec 15 '18 at 12:52
amWhy
1
1
asked Dec 15 '18 at 12:16
Winged Blades of GodricWinged Blades of Godric
665
asked Dec 15 '18 at 12:16
Winged Blades of GodricWinged Blades of Godric
665
asked Dec 15 '18 at 12:16
Winged Blades of GodricWinged Blades of Godric
665
665
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Changing coordinates
$$
x = rcostheta\
y=rsintheta
$$
we get
$$
r^2 (r^2 + 6 cos(2theta)-10) =0
$$
so clearly appear the two solutions
$$
r = 0\
r^2 + 6 cos(2theta)-10=0
$$
answered Dec 15 '18 at 12:38
CesareoCesareo
8,7693516
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1
$begingroup$
This answer should be upvoted
$endgroup$
– Cloud JR
Dec 15 '18 at 12:47
add a comment |
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Not, it's not a bug. Yes, $(0,0)$ belongs to that curve, but it is an isolated point of the curve. In other words, no nearby point belongs to the curve. That's why you can't see it.
answered Dec 15 '18 at 12:18
José Carlos SantosJosé Carlos Santos
160k22127232
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add a comment |
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2 Answers
2
active
oldest
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2 Answers
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active
oldest
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votes
$begingroup$
Changing coordinates
$$
x = rcostheta\
y=rsintheta
$$
we get
$$
r^2 (r^2 + 6 cos(2theta)-10) =0
$$
so clearly appear the two solutions
$$
r = 0\
r^2 + 6 cos(2theta)-10=0
$$
answered Dec 15 '18 at 12:38
CesareoCesareo
8,7693516
$endgroup$
1
$begingroup$
This answer should be upvoted
$endgroup$
– Cloud JR
Dec 15 '18 at 12:47
add a comment |
$begingroup$
Changing coordinates
$$
x = rcostheta\
y=rsintheta
$$
we get
$$
r^2 (r^2 + 6 cos(2theta)-10) =0
$$
so clearly appear the two solutions
$$
r = 0\
r^2 + 6 cos(2theta)-10=0
$$
answered Dec 15 '18 at 12:38
CesareoCesareo
8,7693516
$endgroup$
1
$begingroup$
This answer should be upvoted
$endgroup$
– Cloud JR
Dec 15 '18 at 12:47
add a comment |
$begingroup$
Changing coordinates
$$
x = rcostheta\
y=rsintheta
$$
we get
$$
r^2 (r^2 + 6 cos(2theta)-10) =0
$$
so clearly appear the two solutions
$$
r = 0\
r^2 + 6 cos(2theta)-10=0
$$
answered Dec 15 '18 at 12:38
CesareoCesareo
8,7693516
$endgroup$
Changing coordinates
$$
x = rcostheta\
y=rsintheta
$$
we get
$$
r^2 (r^2 + 6 cos(2theta)-10) =0
$$
so clearly appear the two solutions
$$
r = 0\
r^2 + 6 cos(2theta)-10=0
$$
answered Dec 15 '18 at 12:38
CesareoCesareo
8,7693516
answered Dec 15 '18 at 12:38
CesareoCesareo
8,7693516
answered Dec 15 '18 at 12:38
CesareoCesareo
8,7693516
answered Dec 15 '18 at 12:38
CesareoCesareo
8,7693516
8,7693516
1
$begingroup$
This answer should be upvoted
$endgroup$
– Cloud JR
Dec 15 '18 at 12:47
add a comment |
1
$begingroup$
This answer should be upvoted
$endgroup$
– Cloud JR
Dec 15 '18 at 12:47
1
1
$begingroup$
This answer should be upvoted
$endgroup$
– Cloud JR
Dec 15 '18 at 12:47
$begingroup$
This answer should be upvoted
$endgroup$
– Cloud JR
Dec 15 '18 at 12:47
add a comment |
$begingroup$
Not, it's not a bug. Yes, $(0,0)$ belongs to that curve, but it is an isolated point of the curve. In other words, no nearby point belongs to the curve. That's why you can't see it.
answered Dec 15 '18 at 12:18
José Carlos SantosJosé Carlos Santos
160k22127232
$endgroup$
add a comment |
$begingroup$
Not, it's not a bug. Yes, $(0,0)$ belongs to that curve, but it is an isolated point of the curve. In other words, no nearby point belongs to the curve. That's why you can't see it.
answered Dec 15 '18 at 12:18
José Carlos SantosJosé Carlos Santos
160k22127232
$endgroup$
add a comment |
$begingroup$
Not, it's not a bug. Yes, $(0,0)$ belongs to that curve, but it is an isolated point of the curve. In other words, no nearby point belongs to the curve. That's why you can't see it.
answered Dec 15 '18 at 12:18
José Carlos SantosJosé Carlos Santos
160k22127232
$endgroup$
Not, it's not a bug. Yes, $(0,0)$ belongs to that curve, but it is an isolated point of the curve. In other words, no nearby point belongs to the curve. That's why you can't see it.
answered Dec 15 '18 at 12:18
José Carlos SantosJosé Carlos Santos
160k22127232
edited Dec 18 '18 at 14:25
answered Dec 15 '18 at 12:18
José Carlos SantosJosé Carlos Santos
160k22127232
answered Dec 15 '18 at 12:18
José Carlos SantosJosé Carlos Santos
160k22127232
answered Dec 15 '18 at 12:18
José Carlos SantosJosé Carlos Santos
160k22127232
160k22127232
add a comment |
add a comment |
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