What is the value of the constant C? The curve $y=Cx^{\frac{1}{5}}$ (where  C is constant) is tangent to the line $y=\frac{x}{20}+\frac{32}{5}\:$ somewhere. What is the value of constant C? 
 A: $$\frac{dy}{dx}=C\frac{x^{-\frac45}}5$$ 
At $x=t^5, y=Ct;$  $$\frac{dy}{dx}=\frac C{5t^4}$$
So, the equation of the tangent at $(t^5,Ct)$  will be $$\frac{y-Ct}{x-t^5}=\frac C{5t^4}$$
$$\iff Cx-y(5t^4)+4Ct^4=0$$ which needs to be identical with $$x-20y+128=0$$
So, we need $$\frac C1=\frac{-5t^4}{-20}=\frac{4Ct^4}{128}$$
Can you find $C$ from here?
A: Find out wheere the curve has the same slope as the given line, and then determine the value of $C$ that makes the curve touch the line.
A: The curve $y_c(x)=C x^{\frac{1}{5}}$ touches the line
$y_l(x)=\frac{1}{20}x+\frac{32}{5}$ at some point $(x_t,y_t)$,
so we have two conditions: 
$y_c(x_t)=y_l(x_t)$
 and $\left.\frac{dy_c}{dx}\right|_{x=x_t}=\left.\frac{dy_l}{dx}\right|_{x=x_t}$:
$$\begin{align}
C x_t^{\frac{1}{5}}&=\frac{1}{20}x_t+\frac{32}{5},\ (1) \\
\frac{1}{5}C x_t^{-\frac{4}{5}}&=\frac{1}{20}. \ (2)
\end{align}$$
If we multiply (2) by $x_t$ we'll get 
another expression for the left hand size of (1):
$C x^{\frac{1}{5}}=\dfrac{5}{20} x_t$. Now it's trivial to find $x_t$
and then $C$:

