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meww

  • 3 years ago

Convert the rectangular equation to a polar equation. y^2-x^2=7sqrt(x^2+y^2)

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  1. tkhunny
    • 3 years ago
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    \(x = r\cos(\theta)\) \(y = r\sin(\theta)\) Substitute and Simplify.

  2. meww
    • 3 years ago
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    i know that.. but what i got was -rcos(2theta) = 7r and it says the answer is r=-7/cos(2theta)

  3. tkhunny
    • 3 years ago
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    Why didn't you say so? If you show your work, we can have a much better conversation much more quickly. \(x^{2} = r^{2}\cos^{2}(\theta)\) \(y^{2} = r^{2}\sin^{2}(\theta)\) \(\sqrt{x^{2} + y^{2}} = r\) And we're left with: \(r(\sin^{2}(x) - cos^{2}(x)) = 7\) Sure enough, exctly as you have it. So, what's the dilemma? Not the form allowed by the exam? Try cos(x) = 1/sec(x)

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