the equation of tangent drawn to the curve y^2-2x^3-4y+8=0 from the point (1,2)is

Diff w.r. to x we get
2ydy/dx-6x^2-4dy/dx=0
=>dy/dx(2y-4)=6x^2
=>dy/dx=3x^2/(y-2)
slope of tangent m=(dy/dx) at (1,2)=infinite
i.e. the tangent to the curve at the point (1,2) is perpendicular to the x axis
i.e. it is parallel to y axis
so we can take eq of the tangent is x=k
it passes through (1.2)
so 1=k
hence eq of curve is x=1 Ans.
Thanks & Regards
Rinkoo Gupta
AskIITians Faculty