Is this locus question false? (1 Viewer)

Ambility · Jul 18, 2015

$\text{The point P(x, y) moves so that its distance AP from the point A(5, 1) is always twice its distance BP from the point B(-1,4).}\\\text{Show that the equation of the locus is: }x^2+y^2+6x-10y+14=0}$

I seem to be getting an answer different to the locus given.

InteGrand · Jul 18, 2015

Ambility said:
$\text{The point P(x, y) moves so that its distance AP from the point A(5, 1) is always twice its distance BP from the point B(-1,4).}\\\text{Show that the equation of the locus is: }x^2+y^2+6x-10y+14=0}$

I seem to be getting an answer different to the locus given.

$$AP=\sqrt{(x-5)^2 + (y-1)^2}$ and $BP=\sqrt{(x+1)^2 + (y-4)^2}$.$

$\therefore AP = 2\cdot BP \Rightarrow \sqrt{(x-5)^2 + (y-1)^2} = 2\sqrt{(x+1)^2 + (y-4)^2}$

$$\Rightarrow (x-5)^2 + (y-1)^2 = 4\left[ (x+1)^2 + (y-4)^2 \right]$ (squaring both sides)$

$\Longleftrightarrow x^2 - 10x + 25 + y^2 -2y + 1 = 4\left[ x^2 + 2x + 1+ y^2 - 8y + 16 \right]$

$\Longleftrightarrow x^2 - 10x + 25 + y^2 -2y + 1 = 4x^2 + 8x + 4+ 4y^2 - 32y + 64$

$\Longleftrightarrow 3x^2 + 18x + 3y^2 - 30y + 42 = 0$

$\Longleftrightarrow x^2 + 6x + y^2 - 10y + 14 = 0$ (dividing through by 3). $\blacksquare$

Ambility · Jul 18, 2015

InteGrand said:
$$AP=\sqrt{(x-5)^2 + (y-1)^2}$ and $BP=\sqrt{(x+1)^2 + (y-4)^2}$.$

$\therefore AP = 2\cdot BP \Rightarrow \sqrt{(x-5)^2 + (y-1)^2} = 2\sqrt{(x+1)^2 + (y-4)^2}$

$$\Rightarrow (x-5)^2 + (y-1)^2 = 4\left[ (x+1)^2 + (y-4)^2 \right]$ (squaring both sides)$

$\Longleftrightarrow x^2 - 10x + 25 + y^2 -2y + 1 = 4\left[ x^2 + 2x + 1+ y^2 - 8y + 16 \right]$

$\Longleftrightarrow x^2 - 10x + 25 + y^2 -2y + 1 = 4x^2 + 8x + 4+ 4y^2 - 32y + 64$

$\Longleftrightarrow 3x^2 + 18x + 3y^2 - 30y + 42 = 0$

$\Longleftrightarrow x^2 + 6x + y^2 - 10y + 14 = 0$ (dividing through by 3). $\blacksquare$

Thank you.

Is this locus question false? (1 Viewer)

Ambility

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InteGrand

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Ambility

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