Let image of (t², 2t) in line x + y + 4 = 0 be (h, k).

 $\therefore \frac{\mathrm{h}-{\mathrm{t}}^2}{1}=\frac{\mathrm{k}-2\mathrm{t}}{1}=\frac{-2({\mathrm{t}}^2+2\mathrm{t}+4)}{1^2+1^2}$

⇒ h – t² = k – 2t = – t² – 2t – 4

⇒ h = – 2t – 4 and  k = – t² – 4

 $\Rightarrow \text{t }=–\left(\frac{\mathrm{h}+4}{2}\right)$

 $\therefore \text{k }+\text{ 4 }=–{\left(\frac{\mathrm{h}+4}{2}\right)}^2$

⇒ (h + 4)² = – 4 (k + 4)

    $\therefore$ Locus is (curve C)

 $\therefore$ (x + 4)² = – 4 (y + 4)

 $\therefore$ For point of intersection with y = – 5

(x + 4)² = – 4 (– 5 + 4) = 4

⇒ x + 4 = ± 2 ⇒ x = – 2 or – 6

 $\therefore$ Distance AB will be 4.