Q14 

14. If the normal to the parabola y² = 4ax to point t₁ cuts the parabola again at point t₂ then prove that ${\text{t}}_2^2\text{ ≥ 8.}$

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{normal} \mathrm{to} \mathrm{proabola} {\mathrm{y}}^2=4\mathrm{ax} \mathrm{at} {\mathrm{t}}_1 \mathrm{meets} \mathrm{at} {\mathrm{t}}_2, \mathrm{then} \\ {\mathrm{t}}_2=-{\mathrm{t}}_1-\frac{2}{{\mathrm{t}}_1} \\ \Rightarrow -{\mathrm{t}}_2={\mathrm{t}}_1+\frac{2}{{\mathrm{t}}_1} \\ \Rightarrow -{\mathrm{t}}_1{\mathrm{t}}_2={{\mathrm{t}}_1}^2+2 \\ \Rightarrow {{\mathrm{t}}_1}^2+{\mathrm{t}}_1{\mathrm{t}}_2+2=0 \mathrm{which} \mathrm{is} \mathrm{quadratic} \mathrm{in} {\mathrm{t}}_1. \mathrm{For} \mathrm{roots} \mathrm{to} \mathrm{be} \mathrm{real}, \\ \mathrm{Discriminant}\ge 0 \\ \Rightarrow {{\mathrm{t}}_2}^2-4\times 1\times 2\ge 0 \\ \Rightarrow {{\mathrm{t}}_2}^2\ge 8 \left[\mathrm{Hence} \mathrm{Proved}\right] \end{gathered}$$


Hope this information will clear your doubts about this topic.

If you have any doubts just ask here on the ask and answer forum and our experts will try to help you out as soon as possible.
Regards