Asked by mathexplorer
How can I solve 3*x^2+y^2=300?
Answers
Answered by
Damon
I do not know what you mean by "solve".
This is an ellipse
x^2/1 + y^2/(sqrt 3)^2 = 10^2
x^2/10^2 + y^2/(10 sqrt 3)^2 = 1
center at(0,0)
total length along x axis = 2 * 10 = 20
total length along y axis = 2 * 10 sqrt 3 = 20 sqrt 3
This is an ellipse
x^2/1 + y^2/(sqrt 3)^2 = 10^2
x^2/10^2 + y^2/(10 sqrt 3)^2 = 1
center at(0,0)
total length along x axis = 2 * 10 = 20
total length along y axis = 2 * 10 sqrt 3 = 20 sqrt 3
Answered by
mathexplorer
i am expressing it as an equation. solve for x and y.
Answered by
Damon
well, solve for y for example.
Since it is an ellipse centered at the origin, we know that for every x we better get a positive and a negative value of y
y^2 = 300 - 3 x^2
y = + or - sqrt (300 - 3 x^2)
Since it is an ellipse centered at the origin, we know that for every x we better get a positive and a negative value of y
y^2 = 300 - 3 x^2
y = + or - sqrt (300 - 3 x^2)
Answered by
mathexplorer
Thank you.
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