Mathematics - Parabola Question with Solution | TestHub

Equation of tangent at $\left(2t^2,4t\right)$ is $\mathrm{ty}=\mathrm{x}+2{\mathrm{t}}^2$ $\because$ It is passing through $(-4,0)$ $\therefore 0=-4+2{\mathrm{t}}^2\Rightarrow \mathrm{t}=\pm \sqrt{2}$ $\begin{array}{r}\therefore {\mathrm{A}}_1=(4,4\sqrt{2})\\ {\mathrm{B}}_1=(4,-4\sqrt{2})\end{array}\}\begin{array}{rl}& {\mathrm{OA}}_1=\sqrt{48}=4\sqrt{3}\\ & {\mathrm{A}}_1{\mathrm{B}}_1=8\sqrt{2}\end{array}$ Equation of altitude of $\Delta A_1{B}_1{C}_1$ drawn from $A_1$ is $y-4\sqrt{2}=\sqrt{2}(x-4)$ $\Rightarrow \sqrt{2}x-y=0$ ...(1) Equation of altitude of $\Delta A_1{B}_1{C}_1$ drawn from $C_1$ is $\mathrm{x}=0$ ...(2) Solving (1) and (2) $\Rightarrow$ orthocentre is $(0,0)$ $\therefore$ correct options are (1), (3)