The equation of the tangent at (2,3) on the curve `y^2=a x^3+b` is `y=4x-5.` Find the values of `aa n dbdot`

Options

A. 3,-5
B. 6,-5
C. 6,15
D. 6,-15

Answer

D

Explanation

Given `y^(2)=ax^(2)+b`…………..i
`implies2y(dy)/(dx)=2ax`
`implies (dy)/(dx)=(ax)/y`
`:.` Slope at `(2,3)=((dy)/(dx))_((2,3))=(2a)/3`
But slope of given tangent is `m=4`
`:.(2a)/3=4impliesa=6`
Since point (2,3) lies on the curve so it satisfies the equation of the curve
`:.(3)^(2)=6(2)^(2)+bimpliesb=-15`