Asked by matt
Discuss the equation
2x^2-y^2+4x+4y-4=0
centrr, transverse axis, vertices foci TWo point s equation of asymptope
x=2cost
y=3cost
0 is less than or equal to t and t is less than or equal to pi. Find rectangular equation
2x^2-y^2+4x+4y-4=0
centrr, transverse axis, vertices foci TWo point s equation of asymptope
x=2cost
y=3cost
0 is less than or equal to t and t is less than or equal to pi. Find rectangular equation
Answers
Answered by
Steve
2x^2-y^2+4x+4y-4=0
2x^2+4x - y^2+4y - 4 = 0
2x^2+4x+2 - y^2+4y-4 -4-2+4 = 0
2(x+1)^2 - (y-2)^2 = 2
(x+1)^2 - (y-2)^2/2 = 1
Now just consult the properties of hyperbolas.
x = 2cost
y = 3cost
First off, I think you have a typo, since that is just 3x=2y
If you meant
x = 2cost
y = 3sint, then you have
x/2 = cost
y/3 = sint
and we all know that cos^2t + sin^2t = 1
2x^2+4x - y^2+4y - 4 = 0
2x^2+4x+2 - y^2+4y-4 -4-2+4 = 0
2(x+1)^2 - (y-2)^2 = 2
(x+1)^2 - (y-2)^2/2 = 1
Now just consult the properties of hyperbolas.
x = 2cost
y = 3cost
First off, I think you have a typo, since that is just 3x=2y
If you meant
x = 2cost
y = 3sint, then you have
x/2 = cost
y/3 = sint
and we all know that cos^2t + sin^2t = 1
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