QJEEAdv24Ma54e5: JEE Advanced 2024 Mathematics - Class 11 JEEAdv24Ma54e5 Solution: Let be three points in the -plane. Suppose that the lines and…

Equation of tangent at (2t $^2$ , 4t) is

$\mathrm{ty}=\mathrm{x}+2 \mathrm{t}^2$

$\because$ It is passing through $(-4,0)$

$\therefore 0=-4+2 \mathrm{t}^2 \Rightarrow \mathrm{t}= \pm \sqrt{2}$

$\left.\begin{array}{rl}\therefore & \mathrm{A}_1=(4,4 \sqrt{2}) \\& \mathrm{B}_1=(4,-4 \sqrt{2})\end{array}\right\} \quad \begin{aligned}& \mathrm{OA}_1=\sqrt{48}=4 \sqrt{3} \\& \mathrm{~A}_1 \mathrm{~B}_1=8 \sqrt{2}\end{aligned}$

Equation of altitude of $\Delta A_1 B_1 C_1$ drawn from $A_1$ is

$\begin{aligned}& y-4 \sqrt{2}=\sqrt{2}(x-4) \\& \Rightarrow \sqrt{2} x-y=0 \quad \text{... (1)}\end{aligned}$

Equation of altitude of $\Delta A_1 B_1 C_1$ drawn from $C_1$ is

$\mathrm{x}=0\quad \text{... (2)}$

Solving (1) and (2) $\Rightarrow$ orthocentre is $(0,0)$

$\therefore$ correct options are (A), (C)