Question

Sol. [Key Concept: $PQ$ for a normal chord is minimal at $t^{2} = 2$; then $PQ = 6\sqrt{3}$ and $Q = (8a,\ \mp \ 4\sqrt{2}\ a)$.]

$$\begin{matrix} (PQ)^{2} & \ = a^{2}\left\lbrack \left( t^{2} - t^{2} \right)^{2} + 4\left( t - t^{'} \right)^{2} \right\rbrack \\ & \ = a^{2}\left\lbrack \left( t + t^{'} \right)^{2} - 4tt^{'} \right\rbrack\left\lbrack \left( t + t^{'} \right)^{2} + 4 \right\rbrack \\ t^{'} & \ = - t - \frac{2}{t} \\ & \ = a^{2}\left\lbrack \frac{4}{t^{2}} + 4\left( t^{2} + 2 \right) \right\rbrack\left\lbrack \frac{4}{t^{2}} + 4 \right\rbrack \\ & \ = 16a^{2}\left( t^{2} + \frac{1}{t^{2}} + 2 \right)\left( \frac{1}{t^{2}} + 1 \right) \\ & \ = 16a^{2}\left\lbrack 3 + t^{2} + \frac{1}{t^{4}} + \frac{3}{t^{2}} \right\rbrack \\ PQ^{2} & \ = 16a^{2}\left( 3 + x + \frac{1}{x^{2}} + \frac{3}{x} \right)\ x > 0 \\ \frac{d}{dx}\left( PQ^{2} \right) & \ = 16a^{2}\left( 1 - \frac{2}{x^{3}} - \frac{3}{x^{2}} \right) \\ & \ = 16a^{2}\frac{\left( x^{3} - 3x - 2 \right)}{x^{3}} \\ & \ = \frac{16a^{2}(x + 1)^{2}(x - 2)}{x^{3}} \end{matrix}$$

$PQ^{2}$ is minimum at $t^{2} = 2$

$$\begin{matrix} (PQ)_{\min} & \ = 4a\sqrt{3 + 2 + \frac{1}{4} + \frac{3}{2}} = 6\sqrt{3}a \\ t^{'} & \ = - \sqrt{2} - \frac{2}{\sqrt{2}} = - 2\sqrt{2} \\ \Rightarrow \ Q & \ \equiv (8a, - 4\sqrt{2}a) \\ \text{~}\text{or}\text{~}\ t^{'} & \ = \sqrt{2} + \frac{2}{\sqrt{2}} = 2\sqrt{2} \\ \Rightarrow \ Q & \ \equiv (8a,4\sqrt{2}a) \end{matrix}$$