How do you convert 4=(5x-3)^2+(y-5)^24=(5x−3)2+(y−5)2 into polar form?

1 Answer
Shiva Prakash M V
Feb 12, 2018

(25cos^2theta+sin^2theta)r^2-(30costheta+10sintheta)r+30=0(25cos2θ+sin2θ)r2−(30cosθ+10sinθ)r+30=0 is the polar form

Explanation:

Given:
4=(5x-3)^2+(y-5)^24=(5x−3)2+(y−5)2
x=rcosthetax=rcosθ
y=rsinthetay=rsinθ
Thus,
4=(5rcostheta-3)^2+(rsintheta-5)^24=(5rcosθ−3)2+(rsinθ−5)2
4=25r^2cos^2theta-30rcostheta+9+r^2sin^2theta-10rsintheta+254=25r2cos2θ−30rcosθ+9+r2sin2θ−10rsinθ+25
(25cos^2theta+sin^2theta)r^2-(30costheta+10sintheta)r+9+25-4=0(25cos2θ+sin2θ)r2−(30cosθ+10sinθ)r+9+25−4=0
(25cos^2theta+sin^2theta)r^2-(30costheta+10sintheta)r+30=0(25cos2θ+sin2θ)r2−(30cosθ+10sinθ)r+30=0 is the polar form

Related questions
See all questions in Converting Between Systems
Impact of this question
1294 views around the world
You can reuse this answer
Creative Commons License