Represent the plane curve by a vector-valued function. \frac{x^{2}}{16}-\frac{y^{2}}{4}=1

Ashley Searcy 2021-11-29 Answered
Represent the plane curve by a vector-valued function. $\frac{{x}^{2}}{16}-\frac{{y}^{2}}{4}=1$
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Step 1
$\frac{{x}^{2}}{16}-\frac{{y}^{2}}{4}=1$
let $x=t$
$\therefore \frac{{t}^{2}}{16}-\frac{{y}^{2}}{4}=1$
$\frac{{y}^{2}}{4}=\frac{{t}^{2}}{16}-1$
$\frac{{y}^{2}}{4}=\frac{{t}^{2}-16}{16}$
${y}^{2}=4×\frac{{t}^{2}-16}{16}$
${y}^{2}=\frac{{t}^{2}-16}{4}$
$y=\sqrt{\frac{{t}^{2}-16}{4}}$
$y=\frac{1}{2}\sqrt{{t}^{2}-16}$
Step 2
let r(t) be a vector feild function
$\therefore r\left(t\right)=\left(t\right)i+\left(\frac{1}{2}\sqrt{{t}^{2}-16}\right)j$