Q. 39 - Calculus

Question

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

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

Step-by-Step Solution

Answer

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

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,

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