Explanation

(a)
Let $P\ \left( at_{1},2{at}_{1} \right),\ \mathcal{Q}\ \left( at_{2}^{2},\ 2at_{2} \right)$ be a focal chord of the parabola $y^{2} = 4ax$
Therefore, the tangents at $P$ and $\mathcal{Q}$ meet at $\lbrack at_{1}t_{2},a\left( t_{1} + t_{2}) \right\rbrack\$
Since, $t_{1}t_{2} = - 1$
$x_{1} = - a$
and $y_{1} = a(t_{1} + t_{2})$
and normal at $P$ and $\mathcal{Q}$, meet at
$\lbrack 2a + a(t_{1}^{2} + t_{2}^{2} - 1),\ a(t_{1} + t_{2})\rbrack\$
$\therefore\ x_{2} = 2a + a(t_{1}^{2} + t_{2}^{2} - 1)$
and $y_{2} = a(t_{1} + t_{2})$
$\therefore\ x_{1}x_{2} + y_{1}y_{2} = - a\ \left\lbrack 2a + a\left( t_{1}^{2} + t_{2}^{2} - 1 \right) \right\rbrack + a^{2}\left( t_{1} + t_{2} \right)^{2}$
$= - 3a^{2}$
Now, $x_{1}x_{2} + y_{1}y_{2} = at_{1}^{2}.at_{2}^{2} + 2at_{1}.2at_{2}$
$= a^{2}\left( t_{1}t_{2} \right)^{2} + 4a^{2}\left( t_{1}t_{2} \right)$
$= a^{2} - 4a^{2} = - 3a^{2}\ \ \ \ (\therefore\ t_{1}t_{2} = - 1)$