Let $a$ and $b$ be nonzero real numbers such that
\[(2 - 7i)(a + bi)\]is pure imaginary.  Find $\frac{a}{b}.$
Expanding $(2 - 7i)(a + bi),$ we get
\[2a - 7ai + 2bi - 7bi^2 = 2a - 7ai + 2bi + 7b.\]Since this number is pure imaginary, the real part $2a + 7b$ is equal to 0.  Hence, $\frac{a}{b} = \boxed{-\frac{7}{2}}.$