Let (α,β) be the point whose distance from the point (0,2) is five times it's distance from the line 3x+4y=1 .
The distance between (α,β) and (0,2) is =(α−0)2+(β−2)2
Given equation of the line 3x+4y=1 can be written as 4x+3y=12 ............(1)
Now distance from the point (α,β) to the line is =∣(4)2+(3)24α+3β−12∣ =∣54α+3β−12∣
Therefore according to the problem, (α−0)2+(β−2)2=5.∣54α+3β−12∣
⟹(α−0)2+(β−2)2=∣4α+3β−12∣
⟹(α−0)2+(β−2)2=∣4α+3β−12∣2 [ Squaring both side]
⟹α2+β2−4β+4=16α2+9β2+144+24αβ−96α−72β
⟹15α2+8β2+24αβ−96α−68β+140=0
Hence required locus of the point (α,β) is 15x2+8y2+24xy−96x−68y+140=0
Which the required solution.
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