Suppose a point $\mathrm{P}$ moves so that $\mathrm{BP}^2-\mathrm{AP}^2=121$, where $A$ and $B$ are $(2,5)$…

Suppose a point $\mathrm{P}$ moves so that $\mathrm{BP}^2-\mathrm{AP}^2=121$, where $A$ and $B$ are $(2,5)$ and $(5,11)$ respectively. Then the locus of $\mathrm{P}$ is a straight line, whose slope is
  1. $1 / 2$
  2. $-2$
  3. $-1 / 2$
  4. $2$

Solution

Let $\mathrm{P}(h, k)$ be the point $B P^2-A P^2=121$ $\begin{aligned} & (11-k)^2+(5-h)^2-(h-2)^2-(k-5)^2=121 \\ & 121+k^2-22 k+25+h^2-10 h-h^2-4+4 h \\ & -k^2-25+10 k=121\end{aligned}$ $\begin{aligned} & -6 h-12 k=4 \\ & h+2 k=-4 / 6 \\ & 2 k=-h-4 / 6 \\ & k=-1 / 2 h-4 / 12 \\ & y=-1 / 2 x-1 / 3 \\ & m=-1 / 2\end{aligned}$

Asked in: AP EAMCET 2022 (05 Jul Shift 2)

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