Find a Cartesian equation for the curve and identify it. r2cos⁡(2θ)=1

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Find a Cartesian equation for the curve and identify it.  r2cos⁡(2θ)=1

Anzante2m · Accepted Answer

Step 1&amp;nbsp;We have:&amp;nbsp;The polar formular2cos⁡(2θ)=1&amp;nbsp;Step 2&amp;nbsp;Consider,&amp;nbsp;r2cos⁡(2θ)=1&amp;nbsp;r2(cos2θ−sin2θ)=1&amp;nbsp;(∵cos⁡(2θ)=cos2θ−sin2θ)&amp;nbsp;The parametric equation is an example.x=rcos⁡θ&amp;nbsp;y=rsin⁡θ&amp;nbsp;The implies,&amp;nbsp;xr=cos⁡θ&amp;nbsp;yr=sin⁡θ&amp;nbsp;Then,&amp;nbsp;(xr)2=cos2θ&amp;nbsp;(yr)2=sin2θ&amp;nbsp;Step 3&amp;nbsp;Substitute&amp;nbsp;(xr)2=cos2θ,(yr)2=sin2θ∈r2(cos2θ−sin2θ)=1&amp;nbsp;r2((xr)2−(yr)2)=1&amp;nbsp;r2(x2r2−y2r2)=1&amp;nbsp;r2(1r2(x2−y2))=1&amp;nbsp;x2−y2=1&amp;nbsp;Thus, the Cartesian equation of&amp;nbsp;r2cos⁡(2θ)=1isx2−y2=1

eninsala06 · Accepted Answer

Step 1&amp;nbsp;r2cos⁡2θ=1&amp;nbsp;Recall that:&amp;nbsp;cos⁡2x=cos2x−sin2x&amp;nbsp;r2(cos2θ−sin2θ)=1&amp;nbsp;Choose multiplication&#039;s distributive property over addition and subtraction.r2cos2θ−r2sin2θ=1&amp;nbsp;Rephrase as follows:(rcos⁡θ)2−(rsin⁡θ)2=1&amp;nbsp;To convert the above equation into the Cartesian form, replace r&amp;nbsp;sin⁡θ&amp;nbsp;with y and r&amp;nbsp;cos⁡θ&amp;nbsp;with x&amp;nbsp;x2−y2=1&amp;nbsp;This is a horizontal parabola centered at the origin with lobes opening to the right/left&amp;nbsp;Answer&amp;nbsp;x2−y2=1

nick1337 · Accepted Answer

Step 1 r2cos⁡(2θ)=1 Given Step 2 cos⁡(2θ)=cos2⁡(θ)−sin2⁡(θ Trig identity cos⁡(2θ) Step 3 r2(cos2⁡(θ)−sin2⁡(θ))=1 Rewritten equation, using substitution from previous line Step 4 r2cos2⁡(θ)−r2sin2⁡(θ))=1 Expansion from previous line Step 5 x2−y2=1 x=rcos⁡(θ) y=rsin⁡(θ) Step 6 Graph of x2−y2=1, which is a hyperbola. x2−y2=1

Find a Cartesian equation for the curve and identify it. r^{2}\cos(2\theta)=1

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