JEE Main 2022 (Online) 25th July Evening ShiftQuestion 1
For $z\in\mathbb{C}$ if the minimum value of $\left(|z-3\sqrt2|+|z-p\sqrt2\,i|\right)$ is $5\sqrt2$, then a value of $p$ is
Complex Numbers
Solution
$|z-3\sqrt2|+|z-p\sqrt2 i|\ge$ distance between $(3\sqrt2,0)$ and $(0,p\sqrt2)$, with equality attainable on the segment. Minimum $=\sqrt{18+2p^2}=5\sqrt2\Rightarrow 18+2p^2=50\Rightarrow p^2=16$, so $p=4$ (among options). Answer: (C).