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The Problem is:

If the two end points of a line segment of length $l$ is moving along two orthogonal straight lines, then find the locus of the point on the line segment which divides it by the ratio of $1:2$.

Here is what I've tried to do so far:

enter image description here

Let $(x,y)$ divides length $l$ by $1:2$. So from the figure we can write the following thing:

$x=\dfrac{x_1+2x_2}{3},\ y=\dfrac{y_1+2y_2}{3}$ and, $l^2=(x_2-x_1)^2+(y_2-y_1)^2$

This implies, $x_1^2+4x_2^2+4x_1x_2=9x^2,\ y_1^2+4y_2^2+4y_1y_2=9y^2$ and

$l^2=x_1^2+x_2^2+y_1^2+y_2^2-2(x_1x_2+y_1y_2)$.

Now to find out the locus we've to find the relation between $x,y$ and $l$; so we've to eliminate $x_1,x_2,y_1,y_2$ from the above equations. I've tried different approach to eliminate $x_1,x_2,y_1,y_2$, but can't help it. How to approach further? or is there any other way of doing it? any kind of help or suggestions will be very much appreciated. Thanks.

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2 Answers 2

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HINT:

First, for the sake of simplicity choose $$x_1=r\cos t, y_1=0,x_2=0,y_2=r\sin t$$ where $r$ is the constant length of the hypotenuse

$$3x=r\cos t,3y=2r\cos t$$

$$(6x)^2+(3y)^2=(2r)^2$$ which is clearly an ellipse.

Now use Translation of axes & Rotation of axes to find suitable coordinates.

Observe that the eccentricity is invariant in transformation of axes.

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  • $\begingroup$ thanks got it ... sir @lab bhattacharjee $\endgroup$
    – k.Vijay
    May 6, 2017 at 6:10
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Let the two orthogonal axes be the $x$- and $y$- axis. Let $(0,y_1)$ be the endpoint of the line segment on the $y$-axis, and $(x_2,0)$ be the endpoint of the segment on the x-axis. Let $(x,y) $ be the point on the line segment that divides it in the ratio $1:2$.

Similar triangles:

$x/x_2 = 1/3$.

$y/y_1 = 2/3$.

Pythagoras : $(x_2)^2 + (y_1)^2 = l^2$.

$9x^2 + (9/4)y^2 = l^2$, an ellipse.

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    $\begingroup$ @k.Vijay. Most welcome. Was puzzled how " simple " the solution turns out to be if you set it up with the x-,y-axes . If you want rotated axes you simply rotate your axes, if you want to translate to another origin, it is fairly easily done. $\endgroup$ May 6, 2017 at 8:03
  • $\begingroup$ yes sir you are right, in the question it was mentioned that "the line is moving along two orthogonal straight lines", that's why i assumed any two arbitrary orthogonal straight lines and messed up. $\endgroup$
    – k.Vijay
    May 6, 2017 at 8:25

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