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

#### Solution

The equation of the given curve is

y^{2}=ax^{3}+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 y^{2}=ax^{3}+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.