Question

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)?

Answer 1

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

Answer 2

Thanks

Answer 3

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

Answer 4

no problem :)