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

Options

A. 0
B. 1
C. `-1`
D. 2

Answer

A

Explanation

We have
`y^(2)=ax^(3)+b`
On differentiating both sides w.r.t. x, we get
`2y(dy)/(dx)=3ax^(2)`
`implies(dy)/(dx)=(3ax^(2))/(2y)`
`implies((dy)/(dx))=(3ax^(2))/(2y)`
`implies((dy)/(dx))_((2,3))=(3a(2)^(2))/(2(3))=2a`
Now, slope of line `y=4x-5` is 4.
`therefore`slope of curve is 2a=4
`impliesa=2`
Now, `y^(2)=ax^(3)+b`
since, (2,3) satisfies the curve `y^(2)=ax^(3)+b`.
`therefore(3)^(2)=2(2)^(3)+b`
`implies b=9-16=-7`
`therefore7a+2b=7(2)+2(-7)=0`