If the line `y=4x-5` touches the curve `y^(2)=ax^(3)+b` at the point (2, 3), show that `7a+2b=0`.

Explanation

`y^(2)=ax^(3)+b`
Differentiating both sides w.r.t. x, we get,
`2y(dy)/(dx)=axx3x^(2)+0`
`:. (dy)/(dx)=(3ax^(2))/(2y)`
`:.((dy)/(dx))_("at "(2, 3))=(3a(2)^(2))/(2(3))=2a`= slope of the tangent at (2, 3)
Since the line `y=4x-5` touches the curve at the point (2, 3).
slope of the tangent at (2, 3) is 4.
`:.2a=4" ":.a=2`
Since (2, 3) lies on the curve `y^(2)=ax^(3)+b`,
`(3)^(2)=a(2)^(3)+b" ":.9=8a+b`
`:.9=8(2)+b" "...[because a=2]`
`:. b= -7`
`:.7a+2b=7(2)+2(-7)=0`.