The equation of tangent drawn to the curve y2-2x3-4y+8=0 from the point (1,2) is

Dear Jayati

y²-2x³-4y+8=0

for slope differentiate the corve

 2ydy/dx -6x² -4dy/dx =0

dydx = 3x² /(y-2)

 at point (1,2)  slope is infinity ,i.e line is parallel yo y axis

 so equation of tangent is  x= 1

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