Answer:
None of these
Explanation:
To convert the parametric curve into Cartesian form,
rewrite the equation for
to make
the subject:
![x=2\sin t](https://img.qammunity.org/2023/formulas/mathematics/high-school/rrr5zk9v9iyjawnw7aoxz42fbz6hr7l2vh.png)
![\implies \sin t=(x)/(2)](https://img.qammunity.org/2023/formulas/mathematics/high-school/b6haei4l0qmez7fmq9mh7oazmccut26fin.png)
Substitute this into the given equation for
:
![\begin{aligned}y & =3 \sin t\\\implies y & = 3 \left((x)/(2)\right)\\y& = (3)/(2)x\end{aligned}](https://img.qammunity.org/2023/formulas/mathematics/high-school/xn0iv447xpljet59d8ibsz60bjwoc9c5zx.png)
Therefore, the Cartesian form of the parametric curve is:
![y=(3)/(2)x](https://img.qammunity.org/2023/formulas/mathematics/high-school/f0bpg98tiy128pu5r85tf8i7212bcn2twr.png)
Further Information
![(x^2)/(2)+(y^2)/(3)=1 \quad \textsf{is the equation of a vertical ellipse}](https://img.qammunity.org/2023/formulas/mathematics/high-school/bfzvbrw523kswf6cgtzqtyopitfclh9ih7.png)
![\textsf{with center (0, 0), co-vertex }\sf √(2), \textsf{ and vertex }√(3)](https://img.qammunity.org/2023/formulas/mathematics/high-school/rb32m8rcj6eaq1m9ec5z1j105tdjk5fyat.png)
![x^2+y^2=6 \quad \textsf{is the equation of a circle}](https://img.qammunity.org/2023/formulas/mathematics/high-school/vzp42tewg6apjckg3ae1dlx7duogtxfk3r.png)
![\textsf{with center (0, 0) and radius }\sf √(6)](https://img.qammunity.org/2023/formulas/mathematics/high-school/8d639phu554krh1n9bzf53vl1la8enmbkb.png)
![3x^2+2y^2=1 \quad \textsf{is the equation of a vertical ellipse}](https://img.qammunity.org/2023/formulas/mathematics/high-school/z86l5g078dps6r7coyziddorj2v7bh3zxy.png)
![\textsf{with center (0, 0), co-vertex }\sf (√(3))/(3), \textsf{ and vertex }(√(2))/(2)](https://img.qammunity.org/2023/formulas/mathematics/high-school/140hezt0hpc3xpuxmxqzdg0tu7qwc2tmo5.png)