SOLUTION: find the equation of the tangent line and normal line to the curve x^2+4xy+y^2=13 at (2,1) f'(x)= 2x+xy+2y=0 f'(x,y)=2(2)+(2)(1)+2(1)=0 =4+2+2=8 slope=8 y-y1=m(x-x1) y-1=

Question 1024333: find the equation of the tangent line and normal line to the curve x^2+4xy+y^2=13 at (2,1)
f'(x)= 2x+xy+2y=0
f'(x,y)=2(2)+(2)(1)+2(1)=0
=4+2+2=8 slope=8
y-y1=m(x-x1)
y-1=8(x-2)
y=8x-15 is this correct?
Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!
No, your derivative is incorrect but your logic is correct.
It's a function of x and y so you have to use implicit differentiation.

At (2,1),

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The normal would then be perpendicular to the tangent.
Their slopes are negative reciprocals.

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