On the locus of the point $P(x, y)$ equidistant from $(3,0)$ and $(0,4)$. If $A$ and $B$ are two points that…

On the locus of the point $P(x, y)$ equidistant from $(3,0)$ and $(0,4)$. If $A$ and $B$ are two points that satisfy $4 x=3 y$ and $x=y$ respectively, then the distance between $A$ and $B$ is
  1. $\frac{5}{2}$
  2. $5$
  3. $\frac{25}{4}$
  4. $25$

Solution


$P Q=P R$ $\begin{aligned} & \Rightarrow \quad \sqrt{(x-3)^2+(y-0)^2}=\sqrt{(x-0)^2+(y-4)^2} \\ & \Rightarrow x^2-6 x+9+y^2=x^2+y^2-8 y+16 \\ & \Rightarrow 6 x-8 y+7=0 \\ & \Rightarrow y=\frac{6 x+7}{8}\end{aligned}$ Let $A=\left(\alpha, \frac{6 \alpha+7}{8}\right)$ and $B=\left(\beta, \frac{6 \beta+7}{8}\right)$ According to question, $4 \alpha=3\left(\frac{6 \alpha+7}{8}\right)$ and $\beta=\frac{6 \beta+7}{8}$ $\Rightarrow \quad \alpha=\frac{3}{2}$ and $\beta=\frac{7}{2}$ $\begin{aligned} & \therefore \quad A=\left(\frac{3}{2}, 2\right) \text { and } B=\left(\frac{7}{2}, \frac{7}{2}\right) \\ & \therefore A B=\sqrt{\left(\frac{3}{2}-\frac{7}{2}\right)^2+\left(2-\frac{7}{2}\right)^2}=\frac{5}{2}\end{aligned}$

Asked in: AP EAMCET 2022 (07 Jul Shift 1)

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