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Q. 39

Expert-verified
Found in: Page 772

### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

# Use Cartesian coordinates to express the equations for the hyperbolas determined by the conditions specified in Exercises 38–43. $\text{foci}\left(±6,0\right)\text{, directrices}x=±1$

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

See the step by step solution

## Step 1. Given information.

The given values are,

$\text{foci}\left(±6,0\right)\text{, directrices}x=±1$

## Step 2. Value of variables.

Now,

$\text{The focus points are}\left(6,0\right),\left(-6,0\right)\text{.}\phantom{\rule{0ex}{0ex}}\text{Center}=\left(\frac{6-6}{2},\frac{0+0}{2}\right)\left[\text{since mid point}=\left(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2}\right)\right]\phantom{\rule{0ex}{0ex}}\text{Center}=\left(0,0\right)\phantom{\rule{0ex}{0ex}}\text{Given driectries are}x=±1\text{.}\phantom{\rule{0ex}{0ex}}a=e\phantom{\rule{0ex}{0ex}}\text{Then}ae=6\phantom{\rule{0ex}{0ex}}{a}^{2}=6\phantom{\rule{0ex}{0ex}}c=\sqrt{\left(0-6{\right)}^{2}+\left(0-0{\right)}^{2}}\left[sinceD=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}\right]\phantom{\rule{0ex}{0ex}}c=6\phantom{\rule{0ex}{0ex}}\text{For a hyperbola,}{a}^{2}+{b}^{2}={c}^{2}\phantom{\rule{0ex}{0ex}}6+{b}^{2}={6}^{2}\phantom{\rule{0ex}{0ex}}{b}^{2}=30$

## Step 3. Substitution.

Now, substitute the obtained values in,

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