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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)?
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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0slope 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 pointslope form, you now have the equation of the normal line

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0your welcome., please, i would appreciate a fan for the moment.. :)
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