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
$$\begin{gathered} y^2=a{x}^3+b...\left(1\right) \\ 2y\frac{dy}{dx}=3a{x}^2 \\ \Rightarrow \frac{dy}{dx}=\frac{3a{x}^2}{2y} \\ \text{Slope of tangent=}{\left(\frac{dy}{dx}\right)}_{\left(2,3\right)}\text{=}\frac{12a}{6}\text{=2a} \\ \text{Given that} \\ \text{Slope of tangent= slope of given line} \\ 2a=4 \\ \Rightarrow a=2 \\ \text{Substituting this and}\mathit{\text{x}}\text{= 2,}\mathit{\text{y}}\text{= 3 in (1), we get} \\ 9=16+b \\ \Rightarrow b=-7 \\ \mathrm{Hence}\text{,}a=2\text{and}b=-7 \end{gathered}$$