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Q. 40

Expert-verifiedFound in: Page 725

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.

$\text{foci}(0,\pm 4)\text{, directrices}y=\pm 1$

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

We are given,

$\text{foci}(0,\pm 4)\text{, directrices}y=\pm 1$

Now, as given,

$\text{The focus points are}(0,4),(0,-4)\text{.}\phantom{\rule{0ex}{0ex}}\text{Center}=\left(\frac{0+0}{2},\frac{4-4}{2}\right)\left[\text{since mid point}=\left(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2}\right)\right]\phantom{\rule{0ex}{0ex}}\text{Center}=(0,0)\phantom{\rule{0ex}{0ex}}\text{Given directries are}y=\pm 1\text{.}\phantom{\rule{0ex}{0ex}}\text{That means}\frac{b}{e}=1\phantom{\rule{0ex}{0ex}}b=e\phantom{\rule{0ex}{0ex}}\text{Then}be=4\phantom{\rule{0ex}{0ex}}b\xb7b=4[\text{since}e=b]\phantom{\rule{0ex}{0ex}}{b}^{2}=4\phantom{\rule{0ex}{0ex}}c=\sqrt{(0-4{)}^{2}+(0-0{)}^{2}}\left[\text{since}D=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}\right]\phantom{\rule{0ex}{0ex}}\text{That is the distance from}(0,4)(0,0)\phantom{\rule{0ex}{0ex}}c=4\phantom{\rule{0ex}{0ex}}\text{For a hyperbola,}{a}^{2}+{b}^{2}={c}^{2}\phantom{\rule{0ex}{0ex}}{a}^{2}+4={4}^{2}\phantom{\rule{0ex}{0ex}}{a}^{2}=12$

Now, substitute the obtained values,

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

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