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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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Answer:

x^2+3xy+y^2=5. at (1,1)2x+3y+3xdy/dx+2ydy/dx=0dy/dx(3x+2y)=-(2x+3y)dy/dx=-(2x+3y)÷(3x+2y)dy/dx(1,1)=-(2+3)÷(3+2)dy/dx=-1eq of the tangent(x-1)=-1(y-1)(x-1)=-y

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