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 ?

Answer 1

The slope of the given line y = 4x − 5 is 4

\[y^2 = a x^3 + b . . . \left( 1 \right)\]

\[2y \frac{dy}{dx} = 3a x^2 \]

\[ \Rightarrow \frac{dy}{dx} = \frac{3a x^2}{2y}\]

\[\text { Slope of tangent }= \left( \frac{dy}{dx} \right)_\left( 2, 3 \right) =\frac{12a}{6}=2a\]

\[\text { Given that }\]

\[\text { Slope of tangent = slope of given line }\]

\[2a = 4\]

\[ \Rightarrow a = 2\]

\[\text { Substituting this and }x= 2,y= 3 \text{ in (1), we get }\]

\[9 = 16 + b\]

\[ \Rightarrow b = - 7\]

\[\text { Hence, a}= 2 \text { and }b = - 7\]