Question

c6 - parametric equations: problem 7
(1 point)
the ellipse
\\( \frac { x ^ { 2 } } { 2 ^ { 2 } } + \frac { y ^ { 2 } } { 4 ^ { 2 } } = 1 \\)
can be drawn with parametric equations. assume the curve is traced clockwise as the parameter increases.
if \\( x = 2 \cos ( t ) \\)
then \\( y = \\)

Explanation:

Step1: Substitute \(x = 2\cos(t)\) into the ellipse equation

Substitute \(x\) into \(\frac{x^{2}}{2^{2}}+\frac{y^{2}}{4^{2}} = 1\). We get \(\frac{(2\cos(t))^{2}}{2^{2}}+\frac{y^{2}}{4^{2}}=1\).
Simplify \(\frac{(2\cos(t))^{2}}{2^{2}}\): \(\frac{4\cos^{2}(t)}{4}=\cos^{2}(t)\). So the equation becomes \(\cos^{2}(t)+\frac{y^{2}}{16}=1\).

Step2: Solve for \(y\)

Rearrange the equation \(\cos^{2}(t)+\frac{y^{2}}{16}=1\) to isolate \(y\).
First, \(\frac{y^{2}}{16}=1 - \cos^{2}(t)\).
Since \(1-\cos^{2}(t)=\sin^{2}(t)\), then \(y^{2}=16\sin^{2}(t)\).
Take the square - root of both sides: \(y=\pm4\sin(t)\).
Because the curve is traced clockwise as \(t\) increases. For the standard parametric equations of an ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = 1\) (\(a = 2\), \(b = 4\)), the clock - wise orientation gives \(y=-4\sin(t)\) (using the right - hand rule for parametric curves).

Answer:

\(y=-4\sin(t)\)