Answer to Question #155299 in Analytic Geometry for Sourav Mondal

Question

Find the locus of the point whose distance from 

the point (0, 2) is five times its distance from the line x/3+y/4=1

Answer 1

Let $(\alpha,\beta)$ be the point whose distance from the point $(0,2)$ is five times it's distance from the line $\frac {x}{3}+\frac{y}{4}=1$ .

The distance between $(\alpha,\beta)$ and $(0,2)$ is $=\sqrt{(\alpha-0)^2+(\beta-2)^2}$

Given equation of the line $\frac {x}{3}+\frac{y}{4}=1$ can be written as $4x+3y=12$ ............(1)

Now distance from the point $(\alpha,\beta)$ to the line is $=|\frac{4\alpha+3\beta-12}{\sqrt {(4)^2+(3)^2}}|$ $=|\frac{4\alpha+3\beta-12}{5}|$

Therefore according to the problem, $\sqrt{(\alpha-0)^2+(\beta-2)^2}=5.|\frac{4\alpha+3\beta-12}{5}|$

$\implies \sqrt{(\alpha-0)^2+(\beta-2)^2}=|4\alpha+3\beta-12|$

$\implies (\alpha-0)^2+(\beta-2)^2=|4\alpha+3\beta-12|^2$ [ Squaring both side]

$\implies \alpha^2+\beta^2-4\beta+4=16\alpha^2+9\beta^2+144+24\alpha\beta-96\alpha-72\beta$

$\implies 15\alpha^2+8\beta^2+24\alpha\beta-96\alpha-68\beta+140=0$

Hence required locus of the point $(\alpha,\beta)$ is $15x^2+8y^2+24xy-96x-68y+140=0$

Which the required solution.