Question #155299

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


1
Expert's answer
2021-01-13T19:34:06-0500

Let (α,β)(\alpha,\beta) be the point whose distance from the point (0,2)(0,2) is five times it's distance from the line x3+y4=1\frac {x}{3}+\frac{y}{4}=1 .

The distance between (α,β)(\alpha,\beta) and (0,2)(0,2) is =(α0)2+(β2)2=\sqrt{(\alpha-0)^2+(\beta-2)^2}

Given equation of the line x3+y4=1\frac {x}{3}+\frac{y}{4}=1 can be written as 4x+3y=124x+3y=12 ............(1)

Now distance from the point (α,β)(\alpha,\beta) to the line is =4α+3β12(4)2+(3)2=|\frac{4\alpha+3\beta-12}{\sqrt {(4)^2+(3)^2}}| =4α+3β125=|\frac{4\alpha+3\beta-12}{5}|

Therefore according to the problem, (α0)2+(β2)2=5.4α+3β125\sqrt{(\alpha-0)^2+(\beta-2)^2}=5.|\frac{4\alpha+3\beta-12}{5}|

    (α0)2+(β2)2=4α+3β12\implies \sqrt{(\alpha-0)^2+(\beta-2)^2}=|4\alpha+3\beta-12|

    (α0)2+(β2)2=4α+3β122\implies (\alpha-0)^2+(\beta-2)^2=|4\alpha+3\beta-12|^2 [ Squaring both side]

    α2+β24β+4=16α2+9β2+144+24αβ96α72β\implies \alpha^2+\beta^2-4\beta+4=16\alpha^2+9\beta^2+144+24\alpha\beta-96\alpha-72\beta

    15α2+8β2+24αβ96α68β+140=0\implies 15\alpha^2+8\beta^2+24\alpha\beta-96\alpha-68\beta+140=0

Hence required locus of the point (α,β)(\alpha,\beta) is 15x2+8y2+24xy96x68y+140=015x^2+8y^2+24xy-96x-68y+140=0

Which the required solution.


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