Question

The normal to the curve at `y=asqrtx+b/sqrtx` where a and b are constants is `4x+y=22` at the point where x=4 and how do you find a and b?

Answer 1

`a=2, and, b=4`.

Explanation:

We know that the slope of normal to the curve

`: y = f(x) "at" x=4 "is" -1/(f'(4))`.

We have, `y=f(x)=asqrtx+b/sqrtx`

`rArr dy/dx=f'(x)=a/(2sqrtx)+b(-1/2*x^(-3/2))`

`rArr f'(4)=a/4-b/16=(4a-b)/16`

`:. "the slope of normal at"`x=4`"is" -1/(f'(4))=16/(b-4a)`

But, from the eqn. of normal, the slope is `-4`. hence,

`16/(b-4a)=-4 rArr 4a-b=4..............................................(1)`

When `x=4, y=f(4)=2a+b/2.`

Hence, the pt. of contact is `(4,2a+b/2)`, which lies on the normal

line `: 4x+y=22`, so that, we get,

`16+2a+b/2=22, i.e., 2a+b/2=6, or, 4a+b=12......(2)`.

Solving `(1) & (2), a=2, and, b=4`.

Enjoy Maths.!