If the equation of the tangent to the curve y^2=a x^3+b at the point…

curve: $y^2=ax^3+b$

so at (2, 3)

$⇒ 3^2=a×2^3+b$

so $8a+b=9$  ...(1)

tangent: $y=4x-5$

$Slope]_{at(2,3)}=4$

$⇒y'=4$

So tangent of curve

$2yy'=30x^2$

so $2×3×4=3×a×2^2$

$⇒a=2$

from (i) → $b=-7$