4

An ellipse of major axis $20√3$ and minor axis $20$ slides along the coordinate axes and always remain confined in the 1st quadrant. The locus of the center of ellipse therefore describes an arc of circle. The length of this arc is $\dots$

Attempt:- Let the foci of the required ellipse be $(X1,Y1)$ & $(X2,Y2)$. Now since it slides along the coordinate axes, the axes are tangents to the required ellipse.

Product of perpendiculars from foci to any tangent is $b²$.

Relevant equations:- center is at {$h=(X1+X2)/2$, $k=(Y1+Y2)/2$},

Distance between foci is $\sqrt{(X1-X2)^2 + (Y1-Y2)^2}=4a^2e^2$

$X1X2=Y1Y2=b^2$

Hence locus of the center is of the ellipse is $X^2 + Y^2 = 400$. But how to find the length of the arc?

Mahdi
  • 627
miyagi_do
  • 1,663

3 Answers3

4

The nontrivial thing here is the fact that the arc in question is indeed circular (I didn't know this). Believing that we can take a look at the following figure where the ellipse stays fixed and the angle formed by the positive $x$- and $y$-axes rotates. As the semiaxes of the ellipse are $10\sqrt{3}$ and $10$ the radius of the arc in question is $20$. The black angle in the figure is $30^\circ$, and during the rotation it increases to $60^\circ$. It follows that our arc has length $20\cdot{\pi\over6}={10\pi\over3}$. This arc is covered four times during a full $2\pi$-turn of the ellipse vs. the angle.

enter image description here

Christian Blatter
  • 226,825
3

When the ellipse is perfectly vertical, its axes will be aligned with the co-ordinate axes and the line joining its centre and the origin of co-ordinates will make an angle of $60^\circ$ with the X-axis. This is because the tangent of the angle will be $\tan\theta=\dfrac{10\sqrt 3}{10}=\sqrt 3$.

Similarly, when the ellipse becomes perfectly horizontal the angle will be $30^\circ ,\implies$that the centre travels only $30^\circ$ on the circle you obtained as the locus. The arc length can now be easily obtained.

enter image description here

najayaz
  • 5,459
  • The center travels 30 only. so the length of the arc should be π/6= length of arc/radius of circle. That comes out to be 10π/3, but the answer is 5π/3. – miyagi_do Apr 03 '15 at 06:38
  • 3
    @yasir well then the answer given in your book is wrong. – najayaz Apr 03 '15 at 07:57
  • @najayaz, how to make this sliding ellipse in Geogebra. I need it. – SarGe Aug 16 '20 at 14:15
  • @SarGe it's been 5 years since I was active on this site. Sorry to disappoint you but I'm not sure I even remember the math I did in this answer, much less how to use Geogebra :) – najayaz Aug 16 '20 at 18:29
  • It's all right, however, the GIF helped me a lot. – SarGe Aug 17 '20 at 02:51
1

The minimal possible distance between the center of the ellipse and either of the coordinate axes is equal to the semiminor axis of the ellipse. Pluggin that into your equation makes it an intersection between line and circle. Once you have coordinates for the resulting positions, you can turn them into angles and then compute the arc length based on that.

MvG
  • 42,596