Answer 1

The equation of the given curve is

y²=ax³+b        .....(1) 

Differentiate equation (1) with respect to x, we get

`2y dy/dx=3ax^2`

`dy/dx=(3ax^2)/(2y) `

`(dy/dx)_(2,3)=((3ax^2)/(2y))_(2,3)`

`=> (dy/dx)_(2,3)=2a`

So, equation of tangent at the point (2, 3) is

(y-3)=2a(x-2)

⇒y=2ax−4a+3      .....(2)

But according to question,

Equation of tangent at the point (2,3) is y=4x−5

Both the equation represents the same line, therefore comparing the coefficients of both the line, we have

2a=4⇒a=2 and 3−4a=−5⇒a=2             .....(3)

The point (2, 3) lies on the curve y²=ax³+b, so

(3)²=a(2)³+b

⇒9=8a+b

⇒9=8×2+b        [From (3)]

⇒b=−7

Hence, the values of a and b are 2 and −7, respectively.