Asked by Mark J
                x=4 cost, y= 2 sint on the interval 
0 ≤ t ≤ 2π
what is the Cartesian equation? nd describe the graph giving the (x,y) co-ordinates ... stating the initial point and the terminal point. ?
i found the cartesian equation as
y= square root of 4-x^2/4
plz help .. i know it a complete circle but i cant have "y" as negative .. so ...
            
        0 ≤ t ≤ 2π
what is the Cartesian equation? nd describe the graph giving the (x,y) co-ordinates ... stating the initial point and the terminal point. ?
i found the cartesian equation as
y= square root of 4-x^2/4
plz help .. i know it a complete circle but i cant have "y" as negative .. so ...
Answers
                    Answered by
            Reiny
            
    cos t = x/4 and sin t = y/2
we know that sin^2 t + cos^2 t = 1
x^2/16 + y^2/4 = 1
times 16
x^2 + 4y^2 = 16
this is NOT a circle, but rather an ellipse
I hope you know the properties of this ellipse
a = 4, b=2
vertices along the x-axis are (-4,0) and (4,0)
along the y-axis they are (0,2) and (0,-2)
    
we know that sin^2 t + cos^2 t = 1
x^2/16 + y^2/4 = 1
times 16
x^2 + 4y^2 = 16
this is NOT a circle, but rather an ellipse
I hope you know the properties of this ellipse
a = 4, b=2
vertices along the x-axis are (-4,0) and (4,0)
along the y-axis they are (0,2) and (0,-2)
                    Answered by
            Mark J
            
    what i did is i found the cartesian equation as x=2 sqrt of 4-y^2 nd then i took the table of value nd chose numbers nd substituded but the problem is that in my cartesian equation "x" can not be negative so i only found y-axis as (0,2) and (0,-2) and x-axis as (4,0) so hw can fix this where did i go wrong? did u like find the coordinates from its properties (a = 4, b=2)? 
    
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