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 ...

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)

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)?

To find the Cartesian equation for the given parametric equations, we need to eliminate the parameter "t" and express "y" in terms of "x" only.

Given the parametric equations:
x = 4cos(t)
y = 2sin(t)
with the interval 0 ≤ t ≤ 2π.

To eliminate "t" in terms of "x" and "y", we can start by squaring both equations:
x^2 = 16cos^2(t)
y^2 = 4sin^2(t)

Since cos^2(t) + sin^2(t) = 1 (from the Pythagorean identity), we can rewrite the above equations as:
x^2 = 16(1 - sin^2(t))
y^2 = 4sin^2(t)

Dividing the first equation by 16 gives:
x^2/16 = 1 - sin^2(t)

Substituting 1 - sin^2(t) with cos^2(t) (from the Pythagorean identity) gives:
x^2/16 = cos^2(t)

Now, let's rearrange the second equation to express sin^2(t) in terms of "y":
y^2/4 = sin^2(t)

Dividing the two equations gives:
(x^2/16) / (y^2/4) = cos^2(t) / sin^2(t)

Simplifying:
(x^2/16) / (y^2/4) = cos^2(t) * csc^2(t)
(x^2/16) / (y^2/4) = cot^2(t)

Finally, we can replace cot^2(t) with its equivalent in terms of "x" and "y" as follows:
(x^2/16) / (y^2/4) = (x^2/y^2)

Since we want to express "y" in terms of "x" only, we can multiply both sides of the equation by y^2 and simplify:
x^2 / 16 = x^2

Cross-multiplying:
x^2 = 16x^2

Bringing all terms to one side:
0 = 15x^2

Dividing by 15:
x^2 = 0

Taking the square root on both sides:
x = 0

Therefore, the Cartesian equation for the given parametric equations is x = 0.

Now, let's describe the graph. The Cartesian equation x = 0 represents a vertical line passing through the y-axis at x = 0. The line is parallel to the x-axis and does not depend on the value of y. Hence, it is a vertical line with the equation x = 0.

The coordinates (x, y) on this line will always have x = 0. Therefore, the initial point and the terminal point are both (0, y), where y can take any value in the range of the interval given (0 ≤ t ≤ 2π).

In summary, the graph of the Cartesian equation x = 0 is a vertical line passing through the y-axis, and the (x, y) coordinates on this line have x = 0 and y taking any value in the given interval range.