Question

Sol. [Key Concept: Take the slope-1 tangent $y = x + a$ and the slope-1 normal; the distance between these parallel lines is $\frac{{|c}_{1} - c_{2}|}{\sqrt{2}}$.]

$$\frac{1}{t_{1}} = 1 \Rightarrow t_{1} = 1$$

Equation of tangent at $A$ is

$$\begin{matrix} yt_{1} & & \ = x + at_{1}^{2} \\ \Rightarrow \ y & \ = x + a & \text{(1)} \\ - t_{2} & \ = 1 \Rightarrow t_{2} = - 1 & \text{(1)} \end{matrix}$$

Equation of normal at $B$ is

$$y = - t_{2}x + 2at_{2} + at_{2}^{3}$$

$$\begin{array}{r} \Rightarrow y = x - a - 2a \Rightarrow y = x - 3a\#(2) \end{array}$$

⟂ ar distance between lines (1) and (2) is

$$\frac{(a + 3a)}{\sqrt{1^{2} + 1^{2}}} = \frac{4a}{\sqrt{2}} = 2\sqrt{2}a$$