If the equal ion of the tangent to the curve y2=ax3+b at the point (2, 3) is y=4x−5, then

The equation of the curve is $y^{2} = a x^{3} + b$ .

$∴ 2 y \frac{d y}{d x} = 3 a x^{2} \Rightarrow \frac{d y}{d x} = \frac{3 a x^{2}}{2 y} \Rightarrow {(\frac{d y}{d x})}_{(2, 3)} = 2 a$

The equation of the tangent at (2, 3) is

$y - 3 = 2 a (x - 3) or, 2 a x - y - 6 a + 3 = 0$

But, the equation of the tangent at (2, 3) is $y = 4 x - 5$ .

Comparing these two equations, we get

$2 a = 4$ and $- 6 a + 3 = - 5 \Rightarrow a = 2$

As (2, 3) lies on the curve $y^{2} = a x^{3} + b$ .

$∴ 9 = 8 a + b \Rightarrow 9 = 16 + b \Rightarrow b = - 7$

If the equal ion of the tangent to the curve $y^{2} = a x^{3} + b$ at the point $(2, 3)$ is $y = 4 x - 5$ , then