Find k in a curve equation when equation of a line tangent to curve is given 
The equation of the curve is
$$y=x\left(\frac{k}{\sqrt{x}} - 1\right)$$
Does the problem mean the curve has a slope of zero at $y = 25/4$? The problem asks to find the value of $k$ and equation of line "l" which can be seen in the graph.
 A: Note that the equation of the curve
$$y = x\left(\frac{k}{\sqrt x} - 1\right) \tag 1$$
is undefined at $x=0$. We can change it to the more convenient
$$y = k\sqrt x - x \tag 2$$
which is equivalent, except that it is defined at zero.
As you mention in the comments,
$$y' = \frac{dy}{dx} = \frac{k}{2\sqrt x} - 1 \tag 3$$
We're given that the curve is tangent to the horizontal line
$$y = 25/4 \tag 4$$
We can find where the curve has a horizontal tangent by setting $y'=0$.
$$\frac{k}{2\sqrt x} = 1$$
$$\sqrt x = \frac{k}{2}$$
$$x = \frac{k^2}{4}$$
Now we can plug that $\sqrt x$, $x$, and $y=25/4$ into eqn (2).
$$25/4 = k \frac{k}{2} - \frac{k^2}{4}$$
$$25/4 = \frac{k^2}{2} - \frac{k^2}{4}$$
$$25/4 = \frac{k^2}{4}$$
Thus $k=5$

For part (ii) of the question, we can see from eqn (3) that for large $x$, the slope of the tangent line to the curve approaches -1 because $\frac{k}{2\sqrt x}$ is small. Eg, at $x=10000$, $y'=\frac{1}{200} - 1$.
However, the tangent line is displaced from the line $y = -x$. In fact, at $x = x_0$, the equation of the tangent line is
$$y = \left(\frac{5}{2\sqrt {x_0}}-1\right)x + \frac52\sqrt{x_0}$$
