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Answers without the blur. Sign up and see all textbooks for free! Q. 40

Expert-verified Found in: Page 725 ### 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. $\text{foci}\left(0,±4\right)\text{, directrices}y=±1$

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

See the step by step solution

## Step 1. Given information.

We are given,

$\text{foci}\left(0,±4\right)\text{, directrices}y=±1$

## Step 2. Value of the variables.

Now, as given,

$\text{The focus points are}\left(0,4\right),\left(0,-4\right)\text{.}\phantom{\rule{0ex}{0ex}}\text{Center}=\left(\frac{0+0}{2},\frac{4-4}{2}\right)\left[\text{since mid point}=\left(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2}\right)\right]\phantom{\rule{0ex}{0ex}}\text{Center}=\left(0,0\right)\phantom{\rule{0ex}{0ex}}\text{Given directries are}y=±1\text{.}\phantom{\rule{0ex}{0ex}}\text{That means}\frac{b}{e}=1\phantom{\rule{0ex}{0ex}}b=e\phantom{\rule{0ex}{0ex}}\text{Then}be=4\phantom{\rule{0ex}{0ex}}b·b=4\left[\text{since}e=b\right]\phantom{\rule{0ex}{0ex}}{b}^{2}=4\phantom{\rule{0ex}{0ex}}c=\sqrt{\left(0-4{\right)}^{2}+\left(0-0{\right)}^{2}}\left[\text{since}D=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}\right]\phantom{\rule{0ex}{0ex}}\text{That is the distance from}\left(0,4\right)\left(0,0\right)\phantom{\rule{0ex}{0ex}}c=4\phantom{\rule{0ex}{0ex}}\text{For a hyperbola,}{a}^{2}+{b}^{2}={c}^{2}\phantom{\rule{0ex}{0ex}}{a}^{2}+4={4}^{2}\phantom{\rule{0ex}{0ex}}{a}^{2}=12$

## Step 3. Substitution.

Now, substitute the obtained values,

$\frac{\left(y-k{\right)}^{2}}{{b}^{2}}-\frac{\left(x-h{\right)}^{2}}{{a}^{2}}=1\text{where}\left(h,k\right)\text{is the center.}\phantom{\rule{0ex}{0ex}}\frac{\left(y-0{\right)}^{2}}{4}-\frac{\left(x-0{\right)}^{2}}{12}=1\left[\text{since}{a}^{2}=12,{b}^{2}=4\right]\phantom{\rule{0ex}{0ex}}\frac{{y}^{2}}{4}-\frac{{x}^{2}}{12}=1$ ### Want to see more solutions like these? 