Let $A_1, B_1, C_1$ be three points in the $x y$-plane. Suppose that the lines $A_1 C_1$ and $B_1 C_1$ are…
- The length of the line segment $O A_1$ is $4 \sqrt{3}$
- The length of the line segment $A_1 B_1$ is 16
- The orthocenter of the triangle $A_1 B_1 C_1$ is $(0,0)$
- The orthocenter of the triangle $A_1 B_1 C_1$ is $(1,0)$
Solution
Equation of tangent at $\left(2 t^2, 4 t\right)$ is
$\mathrm{ty}=\mathrm{x}+2 \mathrm{t}^2$
$\because$ It is passing through $(-4,0)$
$\therefore 0=-4+2 \mathrm{t}^2 \Rightarrow \mathrm{t}= \pm \sqrt{2}$
$\left.\begin{array}{rl}\therefore \mathrm{A}_1=(4,4 \sqrt{2}) \\ \mathrm{B}_1=(4,-4 \sqrt{2})\end{array}\right\} \begin{aligned} & \mathrm{OA}_1=\sqrt{48}=4 \sqrt{3} \\ & \mathrm{~A}_1 \mathrm{~B}_1=8 \sqrt{2}\end{aligned}$
Equation of altitude of $\Delta A_1 B_1 C_1$ drawn from $A_1$ is
$y-4 \sqrt{2}=\sqrt{2}(x-4)$
$\Rightarrow \sqrt{2} x-y=0$ ...(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)Asked in: JEE Advanced 2024 (Paper 2)