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anonymous

  • 5 years ago

9x^2 + 4y^2 = 36 18x + 8yy' = 0 4yy' = -9x y' = -9x / 4y What's the equation of the normal line at the point (1 , [3sqrt(3)]/2)?

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  1. anonymous
    • 5 years ago
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    slope of the tangent line is: \[m=y'=\left( -9*1 \right)/\left( 4*3\sqrt{3} /2\right)\] \[m=-9/6\sqrt{3}=-3/2\sqrt{3}\] therefore the slope of the normal line is the negative reciprocal of m: \[m _{N}=2\sqrt{3}/3\] Using the point-slope form, you now have the equation of the normal line

  2. anonymous
    • 5 years ago
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    Thanks

  3. anonymous
    • 5 years ago
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    your welcome., please, i would appreciate a fan for the moment.. :)

  4. anonymous
    • 5 years ago
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    no problem :)

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