## 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)?

1. anonymous

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

Thanks

3. anonymous

your welcome., please, i would appreciate a fan for the moment.. :)

4. anonymous

no problem :)