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Answers without the blur. Sign up and see all textbooks for free! Q. 41

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. $\text{foci}\left(0,±4\right)\text{, directrices}y=±2$

The equation is $\frac{{y}^{2}}{8}-\frac{{x}^{2}}{8}=1$.

See the step by step solution

## Step 1. Given information.

The given values are,

$\text{foci}\left(0,±4\right)\text{, directrices}y=±2$

## Step 2. Value of the variables.

Now,

$\text{The focus points are}\left(0,4\right),\left(0,-4\right)\text{.}\phantom{\rule{0ex}{0ex}}\text{Center}=\left(\frac{0+0}{2},\frac{4-4}{2}\right)\left[\text{since mid point}=\left(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2}\right)\right]\phantom{\rule{0ex}{0ex}}=\left(0,0\right)\phantom{\rule{0ex}{0ex}}\text{Given driectries are}y=±2\text{.}\phantom{\rule{0ex}{0ex}}\text{That means}\frac{b}{e}=2\phantom{\rule{0ex}{0ex}}e=\frac{b}{2}\phantom{\rule{0ex}{0ex}}\text{Then}be=4\phantom{\rule{0ex}{0ex}}b·\frac{b}{2}=4\left[sincee=\frac{b}{2}\right]\phantom{\rule{0ex}{0ex}}{b}^{2}=8\phantom{\rule{0ex}{0ex}}c=\sqrt{\left(0-4{\right)}^{2}+\left(0-0{\right)}^{2}}\left[\text{since}D=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}\right]\phantom{\rule{0ex}{0ex}}c=4\phantom{\rule{0ex}{0ex}}\text{For a hyperbola,}{a}^{2}+{b}^{2}={c}^{2}\phantom{\rule{0ex}{0ex}}{a}^{2}+8={4}^{2}\phantom{\rule{0ex}{0ex}}{a}^{2}=8$

## Step 3. Substitution.

On substituting the obtained values,

$\frac{\left(y-k{\right)}^{2}}{{b}^{2}}-\frac{\left(x-h{\right)}^{2}}{{a}^{2}}=1\text{where}\left(h,k\right)\text{is the center.}\phantom{\rule{0ex}{0ex}}\frac{\left(y-0{\right)}^{2}}{8}-\frac{\left(x-0{\right)}^{2}}{8}=1\left[\text{since}{a}^{2}=8,{b}^{2}=8\right]\phantom{\rule{0ex}{0ex}}\frac{{y}^{2}}{8}-\frac{{x}^{2}}{8}=1$ ### Want to see more solutions like these? 