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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,

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