Question

Sol. [Key Concept: For two perpendicular focal chords the quadrilateral area is least when each is inclined at $\frac{\pi}{4}$ to the axis; the minimum equals twice the square on the latus rectum.]

$$\begin{matrix} & P_{1}P_{2} = \left( a + at_{1}^{2} \right) + \left( a + at_{2}^{2} \right) \\ & \ = a\left\lbrack \left( t_{1} + t_{2} \right)^{2} + 4 \right\rbrack\ \left\lbrack \because t_{1}t_{2} = - 1 \right\rbrack \\ & \left. \ Q_{1}Q_{2} = a\left\lbrack t_{3} + t_{4} \right)^{2} + 4 \right\rbrack \\ & \ \frac{2}{\left( t_{1} + t_{2} \right)}\frac{2}{\left( t_{3} + t_{4} \right)} = - 1 \\ & \ \Rightarrow \ \left| \left( t_{1} + t_{2} \right)\left( t_{3} + t_{4} \right) \right| = 4 \end{matrix}$$

Area of $P_{1}Q_{1}P_{2}Q_{2},A = \frac{1}{2}\left| {\overrightarrow{d}}_{1} \times {\overrightarrow{d}}_{2} \right|$

$$\begin{matrix} \left. \ = \frac{1}{2}d_{1}d_{2} = \frac{1}{2}a^{2}\left\lbrack \left( t_{1} + t_{2} \right)^{2} + 4 \right\rbrack\left\lbrack t_{3} + t_{4} \right)^{2} + 4 \right\rbrack \\ A \geq \frac{1}{2}a^{2}2\sqrt{4\left( t_{1} + t_{2} \right)^{2}}2\sqrt{4\left( t_{3} + t_{4} \right)^{2}} \\ = 8a^{2}\left| t_{1} + t_{2} \right|\left| t_{3} + t_{4} \right| = 32a^{2} \end{matrix}$$

at

$$\left( t_{1} + t_{2} \right)^{2} = 4 = \left( t_{3} + t_{4} \right)^{2}$$

$$\Rightarrow t_{1} + t_{2} = 2,t_{3} + t_{4} = - 2 $$or $t_{1} + t_{2} = - 2$,

$$t_{1} + t_{2} = - 2,t_{3} + t_{4} = 2$$

Slope of $P_{1}P_{2} = 1$, slope of $Q_{1}Q_{2} = - 1$ or slope of $P_{1}P_{2} = - 1$, slope of $Q_{1}Q_{2} = 1$