Find the equation of the tangent line to f(x)=1/sqrt x and parallel to x+ky-6=0

Question

Please help, I've been struggling with this for a while.

Answer 1

The line x + ky - 6 = 0 can be solved for y (converted into "y = mx + b" form) as follows:

x + ky - 6 = 0

ky = -x + 6

y = (-1/k)x + 6/k

The slope of this line is -1/k.

The line you are being asked about is parallel to this line, meaning it has the same slope. You also know that the line tangent to f(x) is f'(x).

Since f(x) = x^-1/2, then f'(x) = -1/2x^-3/2 = -1/(2x^3/2)

That means -1/k = -1/(2x^3/2) meaning that k = 2x^3/2

^{Solving this expression for x we get:}

k = 2x^3/2

k/2 = x^3/2

(k/2)^2/3 = x

(k²/4)^1/3 = x

Since y = 1/x^1/2, then, plugging in the expression we have for x we get:

y = 1/((k²/4)^1/3)^1/2

y = 1/(k²/4)^1/6

y = 4^1/6/k^1/3

Using this point ((k²/4)^1/3, 4^1/6/k^1/3) as an (x₁, y₁), we can use the point-slope form of al line to write the equation of the line as:

y - y₁ = m(x - x₁)

y - 4^1/6/k^1/3 = -1/k(x - (k²/4)^1/3)

Note: there are lots of forms of these expressions in "k". If you wanted to rationalize the denominators you could. But this is my view of what seems best.