Question

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

$\text{foci}(3,1)\text{and}(3,9)\text{, directrix}y=4$

Short answer

The equation is$\frac{(y-5{)}^2}{4}-\frac{(x-3{)}^2}{12}=1$.

Step-by-step solution

Step 1. Given information.

It is given that$\text{foci}(3,1)\text{and}(3,9)\text{, directrix}y=4$

Step 2. Value of the variables.

Now,

$$\begin{gathered} \text{The focus points are}(3,1)(3,9)\text{.} \\ \text{Center}=\left(\frac{3+3}{2},\frac{1+9}{2}\right)\left[\text{since mid point}=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\right] \\ \text{Center}=(3,5) \\ \text{Given driectries are}y=4\text{.} \\ \text{That means}\frac{b}{e}=4 \\ \frac{b}{4}=e \\ \text{Then}be=1 \\ {b}^2=4 \\ c=\sqrt{(3-3{)}^2+(5-1{)}^2}\left[\text{since}D=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\right] \\ c=4 \\ \text{For a hyperbola,}{a}^2+b^2=c^2 \\ {a}^2+4=4^2 \\ {a}^2=12 \end{gathered}$$

Step 3. Substitution.

Now, substitute the obtained values,

$$\begin{gathered} \frac{(y-k{)}^2}{b^2}-\frac{(x-h{)}^2}{a^2}=1\text{where}(h,k)\text{is the center.} \\ \frac{(y-5{)}^2}{4}-\frac{(x-3{)}^2}{12}=1\left[\text{since}{a}^2=12,b^2=4\right] \\ \frac{(y-5{)}^2}{4}-\frac{(x-3{)}^2}{12}=1 \end{gathered}$$