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Answers without the blur. Sign up and see all textbooks for free! Q. 32

Expert-verified Found in: Page 1 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # In Exercises 24-34 sketch the parametric curve by eliminating the parameter.$x=\mathrm{sec}t,y=\mathrm{tan}t,t\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$

The graphical representation by using the points$\left(-1,0\right)\left(1,0\right)\left(2,-\sqrt{3}\right)\left(2,\sqrt{3}\right)$ is as follows, Therefore, the equation after elimination of the parameter is ${x}^{2}-{y}^{2}=1$

See the step by step solution

## Step 1: Given information

$x=\mathrm{sec}t,y=\mathrm{tan}t,t\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$

## Step 2: Calculation

Consider the parametric equations $x=\mathrm{sec}t,y=\mathrm{tan}t,t\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right).$

The objective is to sketch the parametric curve by eliminating the parameter.

Take the equation $x=\mathrm{sec}t.$

Square the equation on both sides.

${x}^{2}={\mathrm{sec}}^{2}t$

Take the equation $y=\mathrm{tan}t$.

Square the equation on both sides. Then.

${y}^{2}={\mathrm{tan}}^{2}t$

Now subtract the equation ${y}^{2}={\mathrm{tan}}^{2}t$ from ${x}^{2}={\mathrm{sec}}^{2}t$.

Thus.

${x}^{2}-{y}^{2}={\mathrm{sec}}^{2}t-{\mathrm{tan}}^{2}t\phantom{\rule{0ex}{0ex}}{x}^{2}-{y}^{2}=1\left[\text{since}{\mathrm{sec}}^{2}t-{\mathrm{tan}}^{2}t=1\right]$

In order to draw the graph of the equation assume $x=-1,1,2$.

Substitute $x=-1$ in the equation ${x}^{2}-{y}^{2}=1$.

Then,

${1}^{2}-{y}^{2}=1\phantom{\rule{0ex}{0ex}}1-{y}^{2}=1\phantom{\rule{0ex}{0ex}}y=0\text{simplify}\phantom{\rule{0ex}{0ex}}\left(x,y\right)=\left(-1,0\right)$

Substitute $x=1$ in the equation ${x}^{2}-{y}^{2}=1$

Then.

$\left(-1{\right)}^{2}-{y}^{2}=1\phantom{\rule{0ex}{0ex}}1-{y}^{2}=1\phantom{\rule{0ex}{0ex}}y=0\text{simplify}\phantom{\rule{0ex}{0ex}}\left(x,y\right)=\left(-1,0\right)$

Substitute $x=2$ in the equation ${x}^{2}-{y}^{2}=1$.

Then,

${2}^{2}-{y}^{2}=1\phantom{\rule{0ex}{0ex}}4-{y}^{2}=1$

Add $-4$ on both sides of the equation.

$4-{y}^{2}-4=1-4\phantom{\rule{0ex}{0ex}}4-{y}^{2}-4--3\phantom{\rule{0ex}{0ex}}-{y}^{2}=-3\phantom{\rule{0ex}{0ex}}y=\sqrt{3}\phantom{\rule{0ex}{0ex}}\left(x,y\right)=\left(2,-\sqrt{3}\right)\left(2,\sqrt{3}\right)$

The graphical representation by using the points $\left(-1,0\right)\left(1,0\right)\left(2,-\sqrt{3}\right)\left(2,\sqrt{3}\right)$ is as follows, Therefore, the equation after elimination of the parameter is ${x}^{2}-{y}^{2}=1$ ### Want to see more solutions like these? 