Question

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

$\text{foci}(0,\pm 4)\text{, directrices}y=\pm 2$

Short answer

The equation is$\frac{y^2}{8}-\frac{x^2}{8}=1$.

Step-by-step solution

Step 1. Given information.

The given values are,

$\text{foci}(0,\pm 4)\text{, directrices}y=\pm 2$

Step 2. Value of the variables.

Now,

$$\begin{gathered} \text{The focus points are}(0,4),(0,-4)\text{.} \\ \text{Center}=\left(\frac{0+0}{2},\frac{4-4}{2}\right)\left[\text{since mid point}=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\right] \\ =(0,0) \\ \text{Given driectries are}y=\pm 2\text{.} \\ \text{That means}\frac{b}{e}=2 \\ e=\frac{b}{2} \\ \text{Then}be=4 \\ b·\frac{b}{2}=4\left[sincee=\frac{b}{2}\right] \\ {b}^2=8 \\ c=\sqrt{(0-4{)}^2+(0-0{)}^2}\left[\text{since}D=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\right] \\ c=4 \\ \text{For a hyperbola,}{a}^2+b^2=c^2 \\ {a}^2+8=4^2 \\ {a}^2=8 \end{gathered}$$

Step 3. Substitution.

On substituting the obtained values,

$$\begin{gathered} \frac{(y-k{)}^2}{b^2}-\frac{(x-h{)}^2}{a^2}=1\text{where}(h,k)\text{is the center.} \\ \frac{(y-0{)}^2}{8}-\frac{(x-0{)}^2}{8}=1\left[\text{since}{a}^2=8,b^2=8\right] \\ \frac{y^2}{8}-\frac{x^2}{8}=1 \end{gathered}$$