Question

Use Cartesian coordinates to express the equations for the ellipses determined by the conditions specified in Exercises 32–37.

$\text{foci}\left(0,\pm \frac{3\sqrt{3}}{2}\right)\text{, minor axis}3$

Short answer

The equation is$\frac{4x^2}{9}+\frac{y^2}{9}=1$.

Step-by-step solution

Step 1. Given information.

We are given,

$\text{foci}\left(0,\pm \frac{3\sqrt{3}}{2}\right)\text{, minor axis}3$

Step 2. Value of the variables.

The foci are $\left(0,\frac{3\sqrt{3}}{2}\right)\left(0,-\frac{3\sqrt{3}}{2}\right)$

$$\begin{gathered} \text{center}=\left(\frac{0+0}{2},\frac{\frac{3\sqrt{3}}{2}-\frac{3\sqrt{3}}{2}}{2}\right)\left[\text{since mid point}=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\right] \\ \text{center}=(0,0) \\ \text{Given minor axis is}3. \\ \text{Then}2b=3 \\ b=\frac{3}{2} \\ \mathit{c}\mathbf{=}\sqrt{\mathbf{(}\mathbf{0}\mathbf{-}\mathbf{0}{\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\left(0-\frac{3\sqrt{3}}{2}\right)}^{\mathbf{2}}}\left[\text{since}D=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\right] \\ \text{That is the distance from}\left(0,\frac{3\sqrt{3}}{2}\right)(0,0)\text{.} \\ c=\frac{3\sqrt{3}}{2} \\ \text{Now take}{c}^2=a^2-b^2 \\ {\left(\frac{3\sqrt{3}}{2}\right)}^2=a^2-{\left(\frac{3}{2}\right)}^2 \\ \frac{27}{4}=a^2-\frac{9}{4} \\ {a}^2=9 \end{gathered}$$

Step 3. Substitution.

Now, substitute the obtained values,

$$\begin{gathered} \frac{(x-h{)}^2}{b^2}+\frac{(y-k{)}^2}{a^2}=1\text{where}a>b,(h,k)\text{is the center.} \\ \frac{(x-0{)}^2}{\frac{9}{4}}+\frac{(y-0{)}^2}{9}=1 \\ \frac{4x^2}{9}+\frac{y^2}{9}=1 \end{gathered}$$