I found this question, and I am quoting:
"The line $y=-\frac{x}{5}$ is a normal to the curve $y=x\sqrt{x}-x+c$. Find the value of $c$"
What I don't understand is why didn't they give us the point at which the line $y=-\frac{x}{5}$ is a normal to our curve. Then I believe I could backtrack and find the tangent and integrate the tangent to find the expression for the curve. But right now I don't understand what I am meant to do.
$y = -\dfrac{x}5$ has the same slope throughout, so no matter what point it's normal to the curve at, you know that your curve as a slope of $5$ there. Suppose that the point of intersection if $(x_0,y_0)$. Then you'll first need to solve \begin{align} 5 = \frac{d}{dx}(x\sqrt{x} - x + c)|_{x = x_0} \end{align} For $x_0$. After that, plug into $y = -\dfrac{x}5$ for $y_0$, and get $c$ by substituting $(x,y) = (x_0,y_0)$ into $y = x\sqrt{x} - x + c$