Question

Consider the curve y=ax^1/2 + bx^-1/2 where a and b are constants. the normal to this curve at the point where x=4 is 4x+y=22. Find the values of a and b?

Answer 1

`{(a=2),(b=4):}`

Explanation:

Consider:

`f(x) = ax^(1/2)+bx^(-1/2)`

`f'(x) = 1/2ax^(-1/2) -1/2bx^(-3/2)`

so that:

`f(4) = 2a+b/2 = (4a+b)/2`

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

The equation of the tangent to the curve `y=f(x)` in the point for `x=4` is then:

`y=f(4)-1/(f'(4))(x-4)`

`y=(4a+b)/2 - 16/(4a-b)(x-4)`

`y= - 16/(4a-b)x + ((4a+b)/2 +64/(4a-b))`

`y= - 16/(4a-b)x + (16a^2-b^2+128)/(2(4a-b))`

`16/(4a-b)x + y = (16a^2-b^2+128)/(2(4a-b))`

Compare this to the given equation of the normal line:

`4x+y=22`

and equate the coefficients:

`16/(4a-b) = 4`

`(16a^2-b^2+128)/(2(4a-b) ) = 22`

Start from the first:

`16= 16a-4b`

`16a = 4b+16`

`a= b/4+1`

and substitute in the second:

`(16(b/4+1)^2-b^2+128)/(2(4(b/4+1)-b ) )= 22`

`((b+4)^2-b^2+128)/(2(b+4-b )) = 22`

`(b^2 +8b + 16 -b^2+128)/8 = 22`

`(8b +144)/8 = 22`

`b+18 = 22`

`b=4`

`a=2`