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

Please feel free to post as many doubts on our discussion forum as you can.
 If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very

quickly.
 We are all IITians and here to help you in your IIT JEE  & AIEEE preparation.

 All the best.
 
Regards,
Askiitians Experts
Badiuddin

Approved