ahredent89
2022-08-04
Answered

Find the corresponding rectangular equation represented by the parameteric equations; $x=1+\mathrm{sec}\theta \text{}and\text{}y=2+\mathrm{tan}\theta $ by eliminating the parameter.

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Luna Wells

Answered 2022-08-05
Author has **19** answers

it can be written as $x-1=\mathrm{sec}\theta $...(1)

$y-2=\mathrm{tan}\theta $...(2)

squaring on both sides eqns (1) and (2)and subtracting

$(x-1{)}^{2}-(y-2{)}^{2}={\mathrm{sec}}^{2}\theta -{\mathrm{tan}}^{2}\theta $

$\Rightarrow {x}^{2}+1-2x-{y}^{2}-4-4y=1\text{}\text{}\text{}({\mathrm{sec}}^{2}\mathrm{tan}\theta -{\mathrm{tan}}^{2}\theta =1)$

$\Rightarrow {x}^{2}-{y}^{2}-2x-4y-4=0$

$y-2=\mathrm{tan}\theta $...(2)

squaring on both sides eqns (1) and (2)and subtracting

$(x-1{)}^{2}-(y-2{)}^{2}={\mathrm{sec}}^{2}\theta -{\mathrm{tan}}^{2}\theta $

$\Rightarrow {x}^{2}+1-2x-{y}^{2}-4-4y=1\text{}\text{}\text{}({\mathrm{sec}}^{2}\mathrm{tan}\theta -{\mathrm{tan}}^{2}\theta =1)$

$\Rightarrow {x}^{2}-{y}^{2}-2x-4y-4=0$

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