Question

Use Cartesian coordinates to express the equations for the hyperbolas determined by the conditions specified in Exercises 38–43.

$\text{foci}(0,\pm 4)\text{, directrices}y=\pm 1$

Short answer

The equation is$\frac{y^2}{4}-\frac{x^2}{12}=1$.

Step-by-step solution

Step 1. Given information.

We are given,

$\text{foci}(0,\pm 4)\text{, directrices}y=\pm 1$

Step 2. Value of the variables.

Now, as given,

$$\begin{gathered} \text{The focus points are}(0,4),(0,-4)\text{.} \\ \text{Center}=\left(\frac{0+0}{2},\frac{4-4}{2}\right)\left[\text{since mid point}=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\right] \\ \text{Center}=(0,0) \\ \text{Given directries are}y=\pm 1\text{.} \\ \text{That means}\frac{b}{e}=1 \\ b=e \\ \text{Then}be=4 \\ b·b=4[\text{since}e=b] \\ {b}^2=4 \\ c=\sqrt{(0-4{)}^2+(0-0{)}^2}\left[\text{since}D=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\right] \\ \text{That is the distance from}(0,4)(0,0) \\ c=4 \\ \text{For a hyperbola,}{a}^2+b^2=c^2 \\ {a}^2+4=4^2 \\ {a}^2=12 \end{gathered}$$

Step 3. Substitution.

Now, substitute the obtained values,

$$\begin{gathered} \frac{(y-k{)}^2}{b^2}-\frac{(x-h{)}^2}{a^2}=1\text{where}(h,k)\text{is the center.} \\ \frac{(y-0{)}^2}{4}-\frac{(x-0{)}^2}{12}=1\left[\text{since}{a}^2=12,b^2=4\right] \\ \frac{y^2}{4}-\frac{x^2}{12}=1 \end{gathered}$$