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Question
The equation of the tangent at (2, 3) on the curve y2 = ax3 + b is y = 4x − 5. Find the values of a and b ?
Solution
The slope of the given line y = 4x − 5 is 4
\[y^2 = a x^3 + b . . . \left( 1 \right)\]
\[2y \frac{dy}{dx} = 3a x^2 \]
\[ \Rightarrow \frac{dy}{dx} = \frac{3a x^2}{2y}\]
\[\text { Slope of tangent }= \left( \frac{dy}{dx} \right)_\left( 2, 3 \right) =\frac{12a}{6}=2a\]
\[\text { Given that }\]
\[\text { Slope of tangent = slope of given line }\]
\[2a = 4\]
\[ \Rightarrow a = 2\]
\[\text { Substituting this and }x= 2,y= 3 \text{ in (1), we get }\]
\[9 = 16 + b\]
\[ \Rightarrow b = - 7\]
\[\text { Hence, a}= 2 \text { and }b = - 7\]
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