Cambridge 2 unit Locus and Parabola (1 Viewer)

senarillosk · Nov 14, 2015

Need help! Thank you.

This is question 12 from Exercise 11a

A point P(x,y) moves so that its distance from the point K(2,5) is twice its distance from the line x = -1. Draw a diagram and find the equation of the locus.

The answer in the textbook is 3x² - y² +12x + 10y -25 = 0

Would highly appreciate it if someone explained this to me.

kawaiipotato · Nov 14, 2015

It says that the distance from P to K (D_PK) is twice the distance from P to line x = -1 (I'll call it D_PX)
A written expression for this will be:

$D_{PK} = 2D_{PX}$

$$ When it asks for distance from a point to a straight line, it refers to the perpendicular distance. Hence, the distance $ D_{PX} $ is just the magnitude of the difference between the x-value of P and $ -1$

$$ So, $ D_{PX} = | x - ( -1) | = | x+1 | ($absolute values for magnitude$)$

$D_{PK} = \sqrt{(x-2)^2 + (y-5)^2 } ($ Pythagoras Theorem$ )$

$$ So $ \sqrt{(x-2)^2 + (y-5)^2 } = 2|x+1|$

$(x-2)^2 + (y-5)^2 = 4(x+1)^2 ($ squaring both sides$ )$

$Then just expand and simplify$

InteGrand · Nov 14, 2015

kawaiipotato said:
It says that the distance from P to K (D_PK) is twice the distance from P to line x = -1 (I'll call it D_PX)
A written expression for this will be:

$D_{PK} = 2D_{PX}$

$$ When it asks for distance from a point to a straight line, it refers to the perpendicular distance. Hence, the distance $ D_{PX} $ is just the magnitude of the difference of the x-value of P and $ -1$

$$ So, $ D_{PX} = x - ( -1) = x+1$

$D_{PK} = \sqrt{(x-2)^2 + (y-5)^2 }$

$$ So $ \sqrt{(x-2)^2 + (y-5)^2 } = x+1$

$(x-2)^2 + (y-5)^2 = (x+1)^2$

$Then just expand and simplify$

$$The expression for $D_{PX}$ should be $|x+1|$ (as you said, \textsl{magnitude} of the difference)$.$

Cambridge 2 unit Locus and Parabola (1 Viewer)

senarillosk

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kawaiipotato

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InteGrand

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