Question

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

$\text{foci}(\pm \sqrt{5},1)\text{, major axis}6$

Short answer

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

Step-by-step solution

Step 1. Given information.

The given values are ,

$\text{foci}(\pm \sqrt{5},1)\text{, major axis}6$

Step 2. Value of variables.

The foci are $(\sqrt{5},1)(-\sqrt{5},1)$

$$\begin{gathered} \text{center}=\left(\frac{\sqrt{5}-\sqrt{5}}{2},\frac{1+1}{2}\right) \\ =(0,1) \end{gathered}$$

$$\begin{gathered} Given, \\ 2a=6 \\ a=3 \\ c=\sqrt{(0-\sqrt{5}{)}^2+(1-1{)}^2}\left[\text{since}D=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\right] \\ \text{That is the distance from}(\sqrt{5},1)(0,1)\text{.} \\ c=\sqrt{5} \\ {c}^2=a^2-b^2 \\ (\sqrt{5}{)}^2=3^2-b^2 \\ {b}^2=4 \end{gathered}$$

Step 3. Substitution.

On substituting all the obtained values in,

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