Mathematics - Parabola Question with Solution | TestHub

Given parabola is ${\mathrm{y}}^2=\mathrm{4...}(1)\mathrm{x}$ Let $\mathrm{P}\equiv \left({\mathrm{t}}_1^2,2{\mathrm{t}}_1\right)$ and $\mathrm{Q}\equiv \left({\mathrm{t}}_2^2,2{\mathrm{t}}_2\right)$ Slope of $\mathrm{OP}=\frac{2{\mathrm{t}}_1}{{\mathrm{t}}_1^2}=\frac{2}{{\mathrm{t}}_1}$ and slope of $\mathrm{OQ}=\frac{2}{{\mathrm{t}}_2}$ Since $\mathrm{OP}\perp \mathrm{OQ},\therefore \frac{4}{{\mathrm{t}}_1{\mathrm{t}}_2}=-1$ or ${\mathrm{t}}_1{\mathrm{t}}_2=-4\dots (2)$ Let $\mathrm{R}(\mathrm{h},\mathrm{k})$ be the middle point of $\mathrm{PQ},$ then $\mathrm{h}=\frac{{\mathrm{t}}_1^2+{\mathrm{t}}_2^2}{2}\dots (3)$ and $\mathrm{k}={\mathrm{t}}_1+{\mathrm{t}}_2\dots (4)$ From $(4),{\mathrm{k}}^2={\mathrm{t}}_1^2+{\mathrm{t}}_2^2+2{\mathrm{t}}_1{\mathrm{t}}_2=2\mathrm{h}-8[\mathrm{From}$ (2) and (3)] Hence locus of $\mathrm{R}(\mathrm{h},\mathrm{k})$ is ${\mathrm{y}}^2=2\mathrm{x}-8$