Question

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

$\text{foci}(\pm 1,0),\text{major axis}4$

Short answer

The equation is$\frac{x^2}{4}+\frac{y^2}{3}=1.$

Step-by-step solution

Step 1. Given information.

We are given,

$\text{foci}(\pm 1,0),\text{major axis}4$

Step 2. Value of the variables.

The focus points are $(1,0)(-1,0)$.

The center which is the midpoint of the focus points is,

$$\begin{gathered} =\left(\frac{1-1}{2},0\right)\left[\text{since mid point}=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\right] \\ =(0,0) \\ \text{Given major axis is}4. \\ \text{Then}2a=4 \\ a=2 \\ Now, \\ c=\sqrt{(1+0{)}^2+(0-0{)}^2}\left[\text{since}D=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\right] \\ c=1 \\ Now,Again, \\ {c}^2=a^2-b^2 \\ 1=4-b^2 \\ {b}^2=3 \end{gathered}$$

Step 3. Substitution.

On substituting the obtained values,

$$\begin{gathered} \frac{(x-h{)}^2}{a^2}+\frac{(y-k{)}^2}{b^2}=1\text{where}a>b,(h,k)\text{is the center.} \\ \frac{(x-0{)}^2}{4}+\frac{(y-0{)}^2}{3}=1 \\ \frac{x^2}{4}+\frac{y^2}{3}=1 \end{gathered}$$