Convert hyperbola in rectangular form to polar form3y2−16y−x2+16=0.

Question

Mason Knight · Accepted Answer

Step 1Staring with3y2−16y−x2+16=0Put it first in the standard format. Compute the square in y3(y−d83)2−d643−x2+16=03(y−d83)2−x2=d163d916(y−d83)2−d316x2=1Thus the center is at(0,&amp;nbsp;83),&amp;nbsp;a=43&amp;nbsp;and&amp;nbsp;b=43The focal distancec=a2+b2=4d19+d13=d83Hence, the foci are at&amp;nbsp;(0,0)&amp;nbsp;and&amp;nbsp;(0,163)Taking the focus that is at the origin, then&amp;nbsp;x=rcos⁡θ,&amp;nbsp;y=rsin⁡θBack to the original equation3y2−16y−x2+16=0Substituting the polar expressions,3(rsin⁡θ)2−16rsin⁡θ−r2cos2θ+16=0Using&amp;nbsp;cos2θ=1−sin2θ&amp;nbsp;and collecting termsr2(4sin2θ−1)−16rsin⁡θ+16=0From the quadratic formula, and taking the positive rootr=d12(4sin2θ−1)&amp;nbsp;(16sin⁡θ−256sin2θ−(256sin2θ−64))And this simplifies tor=d12(4sin2θ−1)(16sin⁡θ−8)Factoring the denominator and cancelling equal termsr=d42sin⁡θ+1

Convert hyperbola in rectangular form to polar form

Answered question

Answer & Explanation

New Questions in High school geometry