the tangent to thecurve 3xy^2-2x^2y=1 at(1,1) meets the curve again at which point?

Dear Student,
Please find below the solution to the asked query:

We have3xy2-2x2y=1Differentiate3x.2ydydx+y2-22x.y+x2dydx=0Put x=y=132dydx+1-22+dydx=06dydx+3-4-2dydx=04dydx=1dydx=14y-1=14x-14y-4=x-1x-4y+3=0x=3-4yPut this in curve33-4yy2-2x23-4y=1Now solve to get required point.

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