find the equation of tangent and normal to the curve x^2+3xy+y^2=5 at the point (1,1)
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x^2+3xy+y^2=5. at (1,1)
2x+3y+3xdy/dx+2ydy/dx=0
dy/dx(3x+2y)=-(2x+3y)
dy/dx=-(2x+3y)÷(3x+2y)
dy/dx(1,1)=-(2+3)÷(3+2)
dy/dx=-1
eq of the tangent
(x-1)=-1(y-1)
(x-1)=-y+1
X+y=2
eq of normal
(x-1)=1(y-1)
x-1=y-1
x-y=0
2x+3y+3xdy/dx+2ydy/dx=0
dy/dx(3x+2y)=-(2x+3y)
dy/dx=-(2x+3y)÷(3x+2y)
dy/dx(1,1)=-(2+3)÷(3+2)
dy/dx=-1
eq of the tangent
(x-1)=-1(y-1)
(x-1)=-y+1
X+y=2
eq of normal
(x-1)=1(y-1)
x-1=y-1
x-y=0
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