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April 18, 2014

April 18, 2014

Posted by **Mark J** on Wednesday, October 26, 2011 at 4:07pm.

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 ...

- Math1239 -
**Reiny**, Wednesday, October 26, 2011 at 4:53pmcos 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)

- Math1239 -
**Mark J**, Wednesday, October 26, 2011 at 5:17pmwhat 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)?

- Math1239 -
**Mark J**, Wednesday, October 26, 2011 at 5:40pmwhat 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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