finding equation of the normal

Question

Find the equation of the normal to the curve $y= 4/x$ at point where $y=1/2$. Find the coordinates of the point where this normal cuts the x axis. I know that the curve cuts the x axis when $y=0$ and I tried doing that for starters.

Answer 1

Notice, setting $y=\frac{1}{2}$ in the equation $y=\frac{4}{x}$, $$\frac{1}{2}=\frac{4}{x}\implies x=8$$ Thus the normal is drawn at the point $\left(8, \frac{1}{2}\right)$ to the curve.

Now, the slope of tangent to the curve $y=\frac{4}{x}$ $$\frac{dy}{dx}=-\frac{4}{x^2}$$

Now, the slope of the normal at the point $\left(8, \frac{1}{2}\right)$ $$=\frac{-1}{\frac{dy}{dx}}\mathbb|_{\left(8, \frac{1}{2}\right)}=\frac{-1}{-\frac{4}{(8)^2}}=16$$ hence, the equation of the normal at the point $\left(8, \frac{1}{2}\right)$ is given as $$y-\frac{1}{2}=16(x-8)$$ $$\color{red}{32x-2y-255=0}$$ setting $y=0$ in the equation of the normal, one should get $$32x-2(0)-255=0\implies x=\frac{255}{32}$$ hence the point of intersection of the normal with the x-axis is $\color{blue}{\left(\frac{255}{32}, 0\right)}$