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

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)`
`rArr (dy)/(dx)=(3ax^(2))/(2y)`
`rArr ((dy)/(dx))_((2"," 3)) =(3a(2)^(2))/(2(3))=2a`
Now, slope of line `y=4x-5` is 4
`:.` Slope of curve is 2a = 4
`rArr a=2`
Now, `y^(2)=ax^(3)+b`
Since, (2,3) satisfies the curve
`y^(2)=ax^(3)+b`.
`:.(3)^(2)=2(2)^(3)+b`
`rArrb=9-16=-7`
`:.7a+2b=7(2)+2(-7)=0`