Eliminate parameter from $x=-6cos(4t)$, $y=2 sin(4t)$ [closed]
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Eliminate the parameter in
$$x = -6cos(4t),quad y = 2sin(4t)$$
calculus parametric
closed as off-topic by amWhy, amd, Arnaud D., Brahadeesh, Leucippus Nov 23 at 21:35
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up vote
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Eliminate the parameter in
$$x = -6cos(4t),quad y = 2sin(4t)$$
calculus parametric
closed as off-topic by amWhy, amd, Arnaud D., Brahadeesh, Leucippus Nov 23 at 21:35
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, amd, Arnaud D., Brahadeesh, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
$cos^2(4t)+sin^2(4t)=1$
– Tito Eliatron
Nov 23 at 18:16
I have (x/36)^2 + (y/4)^2 = 1
– Alex Brito
Nov 23 at 18:18
What happens with the (4t) inside the sin and cos?
– Alex Brito
Nov 23 at 18:19
@AlexBrito: $sin^2 x+cos^2 x=1$ in general.
– Yadati Kiran
Nov 23 at 18:48
1
This question shows absolutely no effort on your part. This isn’t a homework-solving site.
– amd
Nov 23 at 19:39
add a comment |
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
Eliminate the parameter in
$$x = -6cos(4t),quad y = 2sin(4t)$$
calculus parametric
Eliminate the parameter in
$$x = -6cos(4t),quad y = 2sin(4t)$$
calculus parametric
calculus parametric
edited Nov 23 at 18:17
egreg
177k1484198
177k1484198
asked Nov 23 at 18:11
Alex Brito
14
14
closed as off-topic by amWhy, amd, Arnaud D., Brahadeesh, Leucippus Nov 23 at 21:35
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, amd, Arnaud D., Brahadeesh, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, amd, Arnaud D., Brahadeesh, Leucippus Nov 23 at 21:35
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, amd, Arnaud D., Brahadeesh, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
$cos^2(4t)+sin^2(4t)=1$
– Tito Eliatron
Nov 23 at 18:16
I have (x/36)^2 + (y/4)^2 = 1
– Alex Brito
Nov 23 at 18:18
What happens with the (4t) inside the sin and cos?
– Alex Brito
Nov 23 at 18:19
@AlexBrito: $sin^2 x+cos^2 x=1$ in general.
– Yadati Kiran
Nov 23 at 18:48
1
This question shows absolutely no effort on your part. This isn’t a homework-solving site.
– amd
Nov 23 at 19:39
add a comment |
$cos^2(4t)+sin^2(4t)=1$
– Tito Eliatron
Nov 23 at 18:16
I have (x/36)^2 + (y/4)^2 = 1
– Alex Brito
Nov 23 at 18:18
What happens with the (4t) inside the sin and cos?
– Alex Brito
Nov 23 at 18:19
@AlexBrito: $sin^2 x+cos^2 x=1$ in general.
– Yadati Kiran
Nov 23 at 18:48
1
This question shows absolutely no effort on your part. This isn’t a homework-solving site.
– amd
Nov 23 at 19:39
$cos^2(4t)+sin^2(4t)=1$
– Tito Eliatron
Nov 23 at 18:16
$cos^2(4t)+sin^2(4t)=1$
– Tito Eliatron
Nov 23 at 18:16
I have (x/36)^2 + (y/4)^2 = 1
– Alex Brito
Nov 23 at 18:18
I have (x/36)^2 + (y/4)^2 = 1
– Alex Brito
Nov 23 at 18:18
What happens with the (4t) inside the sin and cos?
– Alex Brito
Nov 23 at 18:19
What happens with the (4t) inside the sin and cos?
– Alex Brito
Nov 23 at 18:19
@AlexBrito: $sin^2 x+cos^2 x=1$ in general.
– Yadati Kiran
Nov 23 at 18:48
@AlexBrito: $sin^2 x+cos^2 x=1$ in general.
– Yadati Kiran
Nov 23 at 18:48
1
1
This question shows absolutely no effort on your part. This isn’t a homework-solving site.
– amd
Nov 23 at 19:39
This question shows absolutely no effort on your part. This isn’t a homework-solving site.
– amd
Nov 23 at 19:39
add a comment |
3 Answers
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$x^2=36cos^2 4t$, $y^2=4sin^2 4t to x^2+9y^2=36$
How will it be different from $(6cos4t,2sin4t)$
– lab bhattacharjee
Nov 23 at 18:31
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$$({xover 6})^2+({yover 2})^2=1to x^2+9y^2=36$$
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$left(dfrac{x}{6}right)^2+left(dfrac{y}{2}right)^2=1impliesdfrac{x^2}{36}+dfrac{y^2}{4}=1$.
(Equation of an ellipse with length of semi-major and semi-minor axes being $6$ and $2$ units respectively).
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
$x^2=36cos^2 4t$, $y^2=4sin^2 4t to x^2+9y^2=36$
How will it be different from $(6cos4t,2sin4t)$
– lab bhattacharjee
Nov 23 at 18:31
add a comment |
up vote
3
down vote
$x^2=36cos^2 4t$, $y^2=4sin^2 4t to x^2+9y^2=36$
How will it be different from $(6cos4t,2sin4t)$
– lab bhattacharjee
Nov 23 at 18:31
add a comment |
up vote
3
down vote
up vote
3
down vote
$x^2=36cos^2 4t$, $y^2=4sin^2 4t to x^2+9y^2=36$
$x^2=36cos^2 4t$, $y^2=4sin^2 4t to x^2+9y^2=36$
answered Nov 23 at 18:16
Vasya
3,3391515
3,3391515
How will it be different from $(6cos4t,2sin4t)$
– lab bhattacharjee
Nov 23 at 18:31
add a comment |
How will it be different from $(6cos4t,2sin4t)$
– lab bhattacharjee
Nov 23 at 18:31
How will it be different from $(6cos4t,2sin4t)$
– lab bhattacharjee
Nov 23 at 18:31
How will it be different from $(6cos4t,2sin4t)$
– lab bhattacharjee
Nov 23 at 18:31
add a comment |
up vote
1
down vote
$$({xover 6})^2+({yover 2})^2=1to x^2+9y^2=36$$
add a comment |
up vote
1
down vote
$$({xover 6})^2+({yover 2})^2=1to x^2+9y^2=36$$
add a comment |
up vote
1
down vote
up vote
1
down vote
$$({xover 6})^2+({yover 2})^2=1to x^2+9y^2=36$$
$$({xover 6})^2+({yover 2})^2=1to x^2+9y^2=36$$
answered Nov 23 at 18:36
Mostafa Ayaz
13.6k3836
13.6k3836
add a comment |
add a comment |
up vote
0
down vote
$left(dfrac{x}{6}right)^2+left(dfrac{y}{2}right)^2=1impliesdfrac{x^2}{36}+dfrac{y^2}{4}=1$.
(Equation of an ellipse with length of semi-major and semi-minor axes being $6$ and $2$ units respectively).
add a comment |
up vote
0
down vote
$left(dfrac{x}{6}right)^2+left(dfrac{y}{2}right)^2=1impliesdfrac{x^2}{36}+dfrac{y^2}{4}=1$.
(Equation of an ellipse with length of semi-major and semi-minor axes being $6$ and $2$ units respectively).
add a comment |
up vote
0
down vote
up vote
0
down vote
$left(dfrac{x}{6}right)^2+left(dfrac{y}{2}right)^2=1impliesdfrac{x^2}{36}+dfrac{y^2}{4}=1$.
(Equation of an ellipse with length of semi-major and semi-minor axes being $6$ and $2$ units respectively).
$left(dfrac{x}{6}right)^2+left(dfrac{y}{2}right)^2=1impliesdfrac{x^2}{36}+dfrac{y^2}{4}=1$.
(Equation of an ellipse with length of semi-major and semi-minor axes being $6$ and $2$ units respectively).
answered Nov 23 at 18:46
Yadati Kiran
1,442518
1,442518
add a comment |
add a comment |
$cos^2(4t)+sin^2(4t)=1$
– Tito Eliatron
Nov 23 at 18:16
I have (x/36)^2 + (y/4)^2 = 1
– Alex Brito
Nov 23 at 18:18
What happens with the (4t) inside the sin and cos?
– Alex Brito
Nov 23 at 18:19
@AlexBrito: $sin^2 x+cos^2 x=1$ in general.
– Yadati Kiran
Nov 23 at 18:48
1
This question shows absolutely no effort on your part. This isn’t a homework-solving site.
– amd
Nov 23 at 19:39