Locus of midpoints

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Locus of midpoints

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Source: Canada 1970, Problem 8

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#1 h May 14, 2006, 11:43 PM • 2 Y

Y by Adventure10, Mango247

Consider all line segments of length 4 with one end-point on the line $y=x$ and the other end-point on the line $y=2x$ . Find the equation of the locus of the midpoints of these line segments.

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yetti

#2 h May 15, 2006, 9:03 AM • 1 Y

Y by Adventure10

Let A = (a, a), B = (b, 2b) be 2 segment endpoints and C = (x, y) the midpoint of AB. From $AB^2 = 16,$

$(a - b)^2 + (a - 2b)^2 = 16$

$5b^2 - 6ab + 2a^2 - 16 = 0$

$b = \frac{3a \pm \sqrt{80 - a^2}}{5}$

$x = \frac{a + b}{2} = \frac{8a \pm \sqrt{80 - a^2}}{10}$

$y = \frac{a + 2b}{2} = \frac{11a \pm 2\sqrt{80 - a^2}}{10}$

Eliminating the arbitrary parameter a,

$2(10x - 8a) = 10y - 11a,\ \ \ a = 2(2x - y)$

Substituting this to either the equation for x or y (say x), we get the locus equation.

$x = \frac{8(2x - y) \pm \sqrt{20 - (2x - y)^2}}{5}$

$(8y - 11x)^2 = 20 - (2x - y)^2$

$25x^2 - 36xy + 13y^2 = 4$

WHICH IS AN ELLIPSE CENTERED AT THE

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#3 h May 15, 2006, 9:03 AM • 1 Y

Y by Adventure10

(4x-3y)^2 + (3x-2y)^2 = 4

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Kunihiko_Chikaya

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Kunihiko_Chikaya

#4 h May 16, 2006, 2:07 AM • 1 Y

Y by Adventure10

ashwinrk_jain wrote:

(4x-3y)^2 + (3x-2y)^2 = 4

You are right.

Here is my solution.

Let $P(s,s)\in y=x,\ Q(t,2t)\in y=2x,$ we have $\overline{PQ}^2=(s-t)^2+(s-2t)^2=16\ \cdots [*]$

Set the midpoint of $PQ$ be $M(X,Y),$ we have $x=\frac{s+t}{2},\ Y=\frac{s+2t}{2}\Longleftrightarrow s=4X-2Y,\ t=-2X+2Y.$

Plugg these into the equation $[*],$ we obytain $(3X-2Y)^2+(4X-3Y)^2=4,$ which is ashwinrk_jan obtained.

Expand this we obtain $25X^2-36XY+13Y^2=4,$ which yetti obtained.

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#5 h May 16, 2006, 9:19 AM • 2 Y

Y by Adventure10, Mango247

yes i have done the same solution.

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