Solutions

a) We have

$$\begin{align*} f'(a)&=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h} \\ &=\lim_{h\to 0}\frac{\frac{1}{a+h}-\frac{1}{a}}{h} \\ &=\lim_{h\to 0}\frac{\frac{a-(a+h)}{(a+h)a}}{h} \\ &=\lim_{h\to 0}\frac{-1}{a^2+ah} \\ &=-\frac{1}{a^2}. \end{align*}$$

b) The tangent line is defined as

$$y=-\frac{1}{a^2}(x-a)+\frac{1}{a}\implies y=-\frac{x}{a^2}+\frac{2}{a}.$$

If it intersects $f$, then

$$\begin{align*} \frac{1}{x} =-\frac{x}{a^2}+\frac{2}{a} &\implies \frac{x^2}{a^2}-\frac{2x}{a}+1=0 \\ &\implies (x-a)^2=0 \\ &\implies x=a. \end{align*}$$