12ydy/dx-6x2-4dy/dx=0
dy/dx(2y-4)=6x2
dy/dx=6x2/(2y-4)
find dy/dx at (1,2)
dy/dx=∞
=>tangent || to y axis
→y=2 is tha required tangent
find the equations of tangents to the curve y2-2x3-4y+8=0 from the point (1,2)
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4 Answers
62take (y-2)=m(x-1)
here the question says y2-2x3-4y+8=0
ie (y-2)2-2x3+4=0
m2(x-1)2-2x3+4=0
Now use the condition for repeated roots...
1Yes but i get a quadratic in m / cubic eq in x.. but that does'nt give the answer
1
24@anupam
tangent is not at the point (1,2)..it passes through that point[1]