Question

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

$\text{foci}(\pm 6,0)\text{, directrices}x=\pm 1$

Short answer

The equation is$\frac{x^2}{6}-\frac{y^2}{30}=1.$

Step-by-step solution

Step 1. Given information.

The given values are,

$\text{foci}(\pm 6,0)\text{, directrices}x=\pm 1$

Step 2. Value of variables.

Now,

$$\begin{gathered} \text{The focus points are}(6,0),(-6,0)\text{.} \\ \text{Center}=\left(\frac{6-6}{2},\frac{0+0}{2}\right)\left[\text{since mid point}=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\right] \\ \text{Center}=(0,0) \\ \text{Given driectries are}x=\pm 1\text{.} \\ a=e \\ \text{Then}ae=6 \\ {a}^2=6 \\ c=\sqrt{(0-6{)}^2+(0-0{)}^2}\left[sinceD=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\right] \\ c=6 \\ \text{For a hyperbola,}{a}^2+b^2=c^2 \\ 6+b^2=6^2 \\ {b}^2=30 \end{gathered}$$

Step 3. Substitution.

Now, substitute the obtained values in,

$$\begin{gathered} \frac{(x-h{)}^2}{a^2}-\frac{(y-k{)}^2}{b^2}=1\text{where}(h,k)\text{is the center.} \\ \frac{(x-0{)}^2}{6}-\frac{(y-0{)}^2}{30}=1\left[\text{since}{a}^2=6,b^2=30\right] \\ \frac{x^2}{6}-\frac{y^2}{30}=1 \end{gathered}$$