Question

Sol. [Key Concept: Match the tangent $y = mx + \frac{a}{m}$ of $y^{2} = 4ax$ to the tangency condition of $x^{2} - y^{2} = a^{2}$; the sign of $a$ changes the number of common tangents.]

Equation of tangent to $y^{2} = 4ax$ is

$$\begin{array}{r} yt\ = x + at^{2} \Rightarrow y = \frac{1}{t}x + at\#(1) \\ \Rightarrow a^{2}t^{2}\ = a^{2}\frac{1}{t^{2}} - a^{2}\#(1) \end{array}$$

for (1) to represent tangent to $x^{2} - y^{2} = a^{2}$

$${\Rightarrow t^{4} + t^{2} - 1 = 0 }{\Rightarrow t^{2} = \frac{\sqrt{5} - 1}{2} \Rightarrow t = \pm \sqrt{\frac{\sqrt{5} - 1}{2}} }$$⇒ There exists two tangents $\forall a \in R - \{ 0\}$