find the point on the curve 4x=y³ when slope of the tangent is 16/3
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Answer:
Slope of tangent to the given curve
y=
4x−3
−1at(x,y)=
dx
dy
⟹
3
2
=
2
4x−3
1×4
−0
⟹4
4x−3
=3×4
⟹
4x−3
=3
⟹4x−3=9
⟹x=
4
12
⟹x=3
Ifx=3theny=
4×3−3
−1=3−1=2
Therefore, required points is (3,2)
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