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Question 537430: how do i solve and graph this parabola x2+y2+6x-8y=11?
how do i solve this y2/4-x2/25=1
how do i solve this equation 4x2+y2=16
how do i solve this (x-1,782,000,000)2/3.42(10)23 + (y-356,400,000)2/1.368(10)22
Found 2 solutions by stanbon, KMST:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! how do i solve and graph this parabola x2+y2+6x-8y=11?
It is not a parabola; it is a circle:
Complete the square on the x and on the y terms:
(x^2+6x+9) + (y^2-8y+16) = -11+9+16
(x+3)^2+ (y-4)^2= 14
Center at (-3,4); radius = sqrt(14)
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how do i solve this y^2/4-x^2/25=1
Hyperbola with center at (0,0) ; opening up and down;
y-intercept at (0,2) and (0,-2)
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how do i solve this equation 4x^2+y^2=16
Ellipse:
x^2/(1/4) + y^2/16 = 1
Center at (0,0); semi-major axis = 4 ; semi-minor axis = 1/2
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Cheers,
Stan H.
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Answer by KMST(5328)  (Show Source):
You can put this solution on YOUR website!  looks like a circle to me because it has a term in 
and a term in  , and both have the same coefficient
Adding  to both sides you get
 --> 
That is the equation of a circle centered at (-3, 4), with radius 6.
 looks like a hyperbola.
It is centered on the origin and symmetrical with respect to both axes, but you can see that no point with  could be on the graph. This curve hates the x-axis; it won't touch it.
The closer it will get to the x-axis is  <--> when  , so vertices are (0,2) and (0,-2).
The lines  and  are asymptotes.
 is the equation of an ellipse centered at the origin. The ends of its axes are easy to find, by making or
I do not know what you meant by
(x-1,782,000,000)2/3.42(10)23 + (y-356,400,000)2/1.368(10)22
Perhaps it was
but I would expect that to be equal to 1 and be followed by a question about the eccentricity of the orbit of some object that travels along an elliptical path described by that equation.
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