The product of two natural numbers is 17. Then, the sum of the reciprocals of their squares is :

Answer: Option C

Let the numbers be a and b
Then,
ab = 17
⇒ a = 1 and b = 17
So,
$$\begin{array}{rl}& =\frac{1}{a^2}+\frac{1}{b^2}\\ & =\frac{a^2+b^2}{a^2{b}^2}\\ & =\frac{1^2+{\left(17\right)}^2}{{\left(1\times 17\right)}^2}\\ & =\frac{290}{289}\end{array}$$