Q. 43 - Calculus

Question

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

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

Step-by-Step Solution

Answer

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

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,

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