Question

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

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

Short answer

The equation is$\frac{x^2}{4{\alpha }^2}+\frac{y^2}{5{\alpha }^2}=1$.

Step-by-step solution

Step 1. Given information.

The given values are,

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

Step 2. Value of all variables.

The foci are $(\alpha ,0)(-\alpha ,0)$

$$\begin{gathered} \text{center}=\left(\frac{\alpha -\alpha }{2},\frac{0+0}{2}\right)\left[since\text{mid point}=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\right] \\ \text{center}=(0,0) \\ \text{Given major axis is}4\alpha \text{.} \\ \text{Then}2a=4\alpha \\ a=2\alpha \\ c=\sqrt{(0-\alpha {)}^2+(0-0{)}^2}\left[\text{since}D=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\right] \\ \text{That is the distance from}(\alpha ,0)(0,0)\text{.} \\ c=\alpha \\ {c}^2=b^2-a^2 \\ {\alpha }^2=b^2-(2\alpha {)}^2 \\ {b}^2=5{\alpha }^2 \end{gathered}$$

Step 3. Substitution.

Substitute the obtained values in,

$$\begin{gathered} \frac{(x-h{)}^2}{a^2}+\frac{(y-k{)}^2}{b^2}=1\text{where}a>b,(h,k)\text{is the center.} \\ \frac{(x-0{)}^2}{4{\alpha }^2}+\frac{(y-0{)}^2}{5{\alpha }^2}=1 \\ \frac{x^2}{4{\alpha }^2}+\frac{y^2}{5{\alpha }^2}=1 \end{gathered}$$