Question

In Exercises 24-34 sketch the parametric curve by eliminating the parameter.

$x=3\cos t,y=4\sin t,t\in [0,2\pi ]$

Short answer

The required equation after eliminating the parameter is$\frac{x^2}{9}+\frac{y^2}{16}=1$ .

Step-by-step solution

Given information

The parametric curve is$x=3\cos t,y=4\sin t,t\in [0,2\pi ]$

Calculation

Consider the parametric equations$x=3\cos t,y=4\sin t,t\in [0,2\pi ]$.

The objective is to sketch the parametric curve by eliminating the parameter.

Take the equation $x=3\cos t$-

Divide the equation by 3 on both sides.

$$\begin{gathered} \frac{x}{3}=\frac{3\cos t}{3} \\ \frac{x}{3}=\cos t \end{gathered}$$

Squaring the equation on both sides,

${\left(\frac{x}{3}\right)}^2={\cos }^{2}t$

$$\begin{gathered} \frac{x^2}{3^2}={\cos }^{2}t \\ \frac{x^2}{9}={\cos }^{2}t \end{gathered}$$

Now take the equation $y=4\sin t$.

Divide the equation by 4 on both sides.

$$\begin{gathered} \frac{y}{4}=\frac{4\sin t}{4} \\ \frac{y}{4}=\frac{4\sin t}{A} \\ \frac{y}{4}=\sin t \end{gathered}$$

Squaring the equation on both sides.

${\left(\frac{y}{4}\right)}^2={\sin }^{2}t$

$$\begin{gathered} \frac{y^2}{4^2}={\sin }^{2}t \\ \frac{y^2}{16}={\sin }^{2}t \end{gathered}$$

Further simplification

Now add the equations $\frac{x^2}{9}={\cos }^{2}t$and $\frac{y^2}{16}={\sin }^{2}t$.

$$\begin{gathered} \frac{x^2}{9}+\frac{y^2}{16}={\cos }^{2}t+{\sin }^{2}t \\ \frac{x^2}{9}+\frac{y^2}{16}=1 \end{gathered}$$

The equation$\frac{x^2}{9}+\frac{y^2}{16}=1$is an ellipse with center $\left(0,0\right)$with major axis at $-4,4$and the minor axis $-3,3$.

Thus, the points are $(-3,0)(3,0)(0,4)(0,-4)$.

The graphical representation by using the points$(-3,0)(3,0)(0,4)(0,-4)$is as follows,

Therefore, the required equation after eliminating the parameter is $\frac{x^2}{9}+\frac{y^2}{16}=1$.