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=

Algebra ->  Trigonometry-basics -> 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=      Log On


   

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) About Me  (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.
2xdx%2B4%28xdy%2Bydx%29%2B2ydy=0
xdx%2B2xdy%2B2ydx%2Bydy=0
%282x%2By%29dy%2B%282y%2Bx%29dx=0
%282x%2By%29dy=-%282y%2Bx%29dx
dy%2Fdx=-%282y%2Bx%29%2F%282x%2By%29
At (2,1),
m=-%282%281%29%2B2%29%2F%282%282%29%2B1%29
m=-%284%2F5%29
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y-1=-%284%2F5%29%28x-2%29
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The normal would then be perpendicular to the tangent.
Their slopes are negative reciprocals.
m%5Bn%5D%2Am%5Bt%5D=-1
m%5Bn%5D=5%2F4
y-1=%285%2F4%29%28x-2%29
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