\(\displaystyle x = t^{2} + 2t + 1\)

\(\displaystyle x = \left ( t + 1\right )^{2}\)

\(\displaystyle \pm \sqrt{x} = t + 1\)

So 

\(\displaystyle \sqrt{x} = t + 1\) or \(\displaystyle -\sqrt{x} = t + 1\)

We are restricting \(\displaystyle t\) to values on \(\displaystyle \left [ -1, 1\right ]\), so \(\displaystyle t + 1\) is nonnegative; we choose 

\(\displaystyle \sqrt{x} = t + 1\).

Also,

\(\displaystyle y = t^{2} - 2t + 1\)

\(\displaystyle y= \left ( t - 1\right )^{2}\)

\(\displaystyle \sqrt{y} = \pm \left ( t - 1\right )\)

So 

\(\displaystyle \sqrt{y} = t- 1\) or \(\displaystyle -\sqrt{y} = t- 1\)

We are restricting \(\displaystyle t\) to values on \(\displaystyle \left [ -1, 1\right ]\), so \(\displaystyle t - 1\) is nonpositive; we choose

\(\displaystyle -\sqrt{y} = t- 1\)

or equivalently,

\(\displaystyle \sqrt{y} = -t+ 1\)

to make \(\displaystyle t - 1\) nonpositive.

Then,

\(\displaystyle \sqrt{x}+ \sqrt{y} = (t + 1) + (-t + 1)\)

and 

\(\displaystyle \sqrt{x}+ \sqrt{y} = 2\)

\(\displaystyle \sqrt{y} = 2 - \sqrt{x}\)

\(\displaystyle y =\left ( 2 - \sqrt{x} \right )^2\)

\(\displaystyle y = 4 - 4\sqrt{x} + \left (\sqrt{x} \right )^{2}\)

\(\displaystyle y = x + 4 - 4\sqrt{x}\)

Rewrite \(\displaystyle x\) using the double-angle formula:

\(\displaystyle x = \cos 2t =1 - 2\sin ^{2} t = 1 - 2y^{2}\)

Then 

\(\displaystyle x = 1 - 2y^{2}\)

\(\displaystyle x + 2y^{2} = 1\)

which is the correct choice.

\(\displaystyle x = 3t\), so 

\(\displaystyle x^{2} = (3t)^{2} = 9t^{2} = \frac{9}{2} \cdot 2t^{2} = \frac{9}{2} y\).

\(\displaystyle \frac{9}{2} y= x^{2}\), so

\(\displaystyle \frac{2}{9}\cdot \frac{9}{2} y = \frac{2}{9}x^{2}\)

\(\displaystyle y = \frac{2}{9}x^{2}\)

\(\displaystyle y = 4t+2 = 4t - 2 + 4 = 2 (2t - 1)+ 4 = 2x + 4\),

so the Cartesian equation is 

\(\displaystyle y = 2x + 4\).

\(\displaystyle x = \ln (t+1)\)

so 

\(\displaystyle e ^{x} = \left [ \ln (t+1) \right ] ^{x}\)

\(\displaystyle e ^{x} = t + 1\)

\(\displaystyle y = t^2 + 2t + 2 = t^2 + 2t + 1 + 1 = (t + 1)^{2}+ 1 = \left (e^{x} \right )^{2} + 1 = e^{2x} + 1\)

Therefore the Cartesian equation is  \(\displaystyle y = e^{2x} + 1\).

\(\displaystyle y = \sinh t = \frac{e ^{2t} -1}{2e ^{t}} = \frac{\left (e ^{ t} \right) ^{2} -1}{2e ^{t}}\)

\(\displaystyle e^{t} = 10 ^{\log e \cdot t} =\left ( 10 ^{ t} \right )^{\log e} = x ^{\log e}\), so

\(\displaystyle y = \frac{\left ( x ^{\log e} \right) ^{2} -1}{2 x ^{\log e}}= \frac{ x ^{2 \log e} -1}{2 x ^{\log e}}\)

This makes the Cartesian equation

\(\displaystyle y = \frac{ x ^{2 \log e} -1}{2 x ^{\log e}}\).

In order to solve this, we must isolate \(\displaystyle \small t\) in both equations. 

\(\displaystyle \small \small x=3t+4\rightarrow t=(x-4)/3\) and 

\(\displaystyle \small y=1/t\rightarrow t=1/y\).

Now we can set the right side of those two equations equal to each other since they both equal \(\displaystyle \small t\).

 \(\displaystyle \small (x-4)/3=1/y\).

By multiplying both sides by \(\displaystyle \small 3y/(x-4)\), we get \(\displaystyle \small y=3/(x-4)\), which is our equation in rectangular form.

Given \(\displaystyle x=3+t\) and  \(\displaystyle y=4t+7\), we can find \(\displaystyle y\) in terms of \(\displaystyle x\) by isolating \(\displaystyle t\) in both equations:

\(\displaystyle x=3+t\rightarrow t=x-3\)

\(\displaystyle y=4t+7\rightarrow t=\frac{y-7}{4}\)

Since both of these transformations equal \(\displaystyle t\), we can set them equal to each other:

\(\displaystyle \frac{y-7}{4}=x-3\)

\(\displaystyle y-7=4(x-3)\)

\(\displaystyle y-7=4x-12\)

\(\displaystyle y=4x-5\)

In order to find \(\displaystyle y\) with respect to \(\displaystyle x\), we first isolate \(\displaystyle t\) in both equations:\(\displaystyle x=7t-5\rightarrow t=\frac{x+5}{7}\)

\(\displaystyle y=2t+9\rightarrow t=\frac{y-9}{2}\)

Since both equations equal \(\displaystyle t\), we can then set them equal to each other and solve for \(\displaystyle y\):

\(\displaystyle \frac{y-9}{2}=\frac{x+5}{7}\)

\(\displaystyle {y-9}=2(\frac{x+5}{7})\)

\(\displaystyle y=\frac{2}{7}({x+5})+9\)

In order to find \(\displaystyle y\) with respect to \(\displaystyle x\), we first isolate \(\displaystyle t\) in both equations:\(\displaystyle x=4t+9\rightarrow t=\frac{x-9}{4}\)

\(\displaystyle y=-t+2\rightarrow t=2-y\)

Since both equations equal \(\displaystyle t\), we can then set them equal to each other and solve for \(\displaystyle y\):

\(\displaystyle 2-y=\frac{x-9}{4}\)

\(\displaystyle y=2-\frac{x-9}{4}\)

\(\displaystyle y=\frac{8}{4}-\frac{x-9}{4}\)

\(\displaystyle y=\frac{8-(x-9)}{4}\)

\(\displaystyle y=\frac{8-(x-9)}{4}\)

\(\displaystyle y=\frac{8-x+9}{4}\)

\(\displaystyle y=\frac{17-x}{4}\)