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