Question

The equation of tangent at $\left( {2,3} \right)$ on the curve ${y^2} = a{x^3} + b$ is $y = 4x - 5$. Find the values of a and b.

Answer 1

Hint:
The solution of the curve at point of contact with the line is equal to the slope of the line.

Complete step by step solution:
Firstly, we will find the slope of the curve at point $\left( {2,3} \right)$
${y^2} = a{x^3} + b$
On differentiating both sides with respect to x:
$2y \times \dfrac{{dy}}{{dx}} = 3a{x^2}$
$\dfrac{{dy}}{{dx}} = \dfrac{{3a{x^2}}}{{2y}}$
Putting values of $x = 2$ and $y = 3$ for finding the slope of the curve at point $\left( {2,3} \right)$:
$\dfrac{{dy}}{{dx}} = \dfrac{{3a \times {{\left( 2 \right)}^2}}}{{2 \times 3}} = 2a$
$\dfrac{{dy}}{{dx}} = 2a.......\left( 1 \right)$
The equation of tangent line is $y = 4x - 5$
$4x - y - 5 = 0$
General expression of a line is $ax + by + c = 0$
And, slope of a line is $\dfrac{{ - b}}{a}$
Therefore, slope of the line $4x - y - 5 = 0$ is $\dfrac{{ - \left( { - 1} \right)}}{4} = \dfrac{1}{4}$
Since, the solution of the curve at point of contact with the line is equal to the slope of the line.
From equation (1), $\dfrac{{dy}}{{dx}} = 2a = \dfrac{1}{4}$
$a = \dfrac{1}{8}$
Now putting values of $x = 2$, $y = 3$and $a = \dfrac{1}{8}$in the equation of the curve ${y^2} = a{x^3} + b$ to get the value of b:
${\left( 3 \right)^2} = \dfrac{1}{8} \times {\left( 2 \right)^3} + b$
$b = 9$
Hence, values of a and b are $\dfrac{1}{8}$ and $9$ respectively.

Note:
In this type of problem we generally start with finding the slope of a point at the point of contact with the line and equate it with the slope of the line itself to get the values of the coefficients.