Question

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

$\text{foci}(0,\pm \alpha )\text{, minor axis}2\alpha$

Short answer

The answer is$\frac{x^2}{{\alpha }^2}+\frac{y^2}{2{\alpha }^2}=1\text{.}$

Step-by-step solution

Step 1. Given information.

The given values are,

$\text{foci}(0,\pm \alpha )\text{, minor axis}2\alpha$.

Step 2. values of the variables.

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

Centre$=\left(\frac{0+0}{2},\frac{\alpha -\alpha }{2}\right)\left[\text{since mid point}=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\right]$$=(0,0)$

$$\begin{gathered} \text{Given minor axis is}2\alpha \text{.} \\ \text{Then}2b=2\alpha \\ b=\alpha \\ \left.c=\sqrt{(0-0{)}^2+(0-\alpha {)}^2}\text{since}D=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\right] \\ \text{That is the distance from}(0,\alpha )(0,0)\text{.} \\ c=\alpha \\ \text{Now take}{c}^2=a^2-b^2 \\ {c}^2=a^2-b^2 \\ {\alpha }^2=a^2-{\alpha }^2 \\ {a}^2=2{\alpha }^2 \end{gathered}$$

Step 3. Substitution.

On substituting,

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