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An ellipse of major and minor axes of length √3 and 1 respectively, slides along the co-ordinate axes and always remains confined in the first quadrant. The locus of the centre of the ellipse will be the arc of a circle the length of which is:

The answer given is 0.52. In the solution it is given that locus of centre is $x^2+y^2=1$ and difference in parametric angles at two extreme positions is π/6. (This is because the two extreme positions are $tan^{-1} (π/6)$ and $tan^{-1} (π/3)$). I am actually having difficulty understanding why locus of centre is $x^2+y^2=1$. I understand that at the two extreme positions(major axis parallel to x axis and y axis respectively) the distance of the centre from the origin is 1 ($1^2 = (√3/2)^2 + (1/2)^2$), but what about the other positions in between these two? I only have an intuition about the fact that the distance of centre from origin remains 1 as ellipse slides, and would welcome some clearer guidance.

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    $\begingroup$ See the animation of another answer here. $\endgroup$ – Ng Chung Tak May 23 '18 at 14:01
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You can look at it in another way. We know that for an ellipse, if two mutually perpendicular tangents are drawn, then the intersection of these tangents lies on director circle $x^2+y^2=a^2+b^2 = 1$. This equation is valid when $C(0,0)$ is centre of ellipse.

Now here, the origin of coordinates $O(0,0)$ is fixed, but the centre $C$ of ellipse is moving relative to it, so it will also lie on same curve, that is $x^2+y^2 = 1$.

For arc length I think your approach is proper.

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    $\begingroup$ Samjoe. Very nice! $\endgroup$ – Peter Szilas May 23 '18 at 8:05

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