Mathematics - Parabola Question with Solution | TestHub

$\begin{array}{rl}& {\mathrm{m}}_{\mathrm{AB}}={\mathrm{m}}_{\mathrm{AD}}\\ & \Rightarrow \frac{2}{{\mathrm{t}}_1+{\mathrm{t}}_2}=\frac{2\mathrm{a}\left({\mathrm{t}}_1-{\mathrm{t}}_3\right)}{{\mathrm{at}}_1^2-\alpha }\\ & \Rightarrow {\mathrm{at}}_1^2-\alpha =\mathrm{a}\left\{{\mathrm{t}}_1^2-{\mathrm{t}}_1{\mathrm{t}}_3+{\mathrm{t}}_1{\mathrm{t}}_2-{\mathrm{t}}_2{\mathrm{t}}_3\right\}\\ & \Rightarrow \alpha =\mathrm{a}\left({\mathrm{t}}_1{\mathrm{t}}_3+{\mathrm{t}}_2{\mathrm{t}}_3-{\mathrm{t}}_1{\mathrm{t}}_2\right)\\ & \mathrm{AM}=\left|2\mathrm{a}\left({\mathrm{t}}_1-{\mathrm{t}}_3\right)\right|,\mathrm{BN}=\left|2\mathrm{a}\left({\mathrm{t}}_2-{\mathrm{t}}_3\right)\right|,\\ & \mathrm{CD}=\left|{\mathrm{at}}_3^2-\alpha \right|\end{array}$ $\begin{array}{rl}& \mathrm{CD}=\left|{\mathrm{at}}_3^2-\mathrm{a}\left({\mathrm{t}}_1{\mathrm{t}}_3+{\mathrm{t}}_2{\mathrm{t}}_3-{\mathrm{t}}_1{\mathrm{t}}_2\right)\right|\\ & =\mathrm{a}\left|{\mathrm{t}}_3^2-{\mathrm{t}}_1{\mathrm{t}}_3-{\mathrm{t}}_2{\mathrm{t}}_3+{\mathrm{t}}_1{\mathrm{t}}_2\right|\\ & =\mathrm{a}\left|{\mathrm{t}}_3\left({\mathrm{t}}_3-{\mathrm{t}}_1\right)-{\mathrm{t}}_2\left({\mathrm{t}}_3-{\mathrm{t}}_1\right)\right|\\ & \mathrm{CD}=\mathrm{a}\left|\left({\mathrm{t}}_3-{\mathrm{t}}_2\right)\left({\mathrm{t}}_3-{\mathrm{t}}_1\right)\right|\\ & {\left(\frac{\mathrm{AM}\cdot \mathrm{BN}}{\mathrm{CD}}\right)}^2={\left\{\frac{2\mathrm{a}\left({\mathrm{t}}_1-{\mathrm{t}}_3\right)\cdot 2\mathrm{a}\left({\mathrm{t}}_2-{\mathrm{t}}_3\right)}{\mathrm{a}\left({\mathrm{t}}_3-{\mathrm{t}}_2\right)\left({\mathrm{t}}_3-{\mathrm{t}}_1\right)}\right\}}^2\\ & 16{\mathrm{a}}^2=16\times \frac{9}{4}=36\end{array}$