Area of Center Locus
An ellipse with semi-major axis and semi-minor axis is sliding in place, such that it is always tangent to the x-axis at the origin, as shown in the animation image below. The center of the ellipse traces a closed locus. Find the area enclosed by this locus in terms of and .
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Standard parametric equation of ellipse is ( a cos t , b sin t ) . Instead of letting center of ellipse move around, we will consider a scenario where center of ellipse is fixed at the origin of coordinate system S and where we set an origin of another coordinate system S' at some point on the ellipse such that its x'-axis coincide with tangent line to the ellipse at that point. So, radius vector of origin of S' with respect to S is ( a cos t , b sin t ) . Unit vector i ′ which corresponds to x'-axis is consequently: i ′ = t ^ = a 2 sin 2 t + b 2 cos 2 t 1 ( − a sin t , b cos t ) Unit vector j ′ which corresponds to y'-axis and is orthogonal to i ′ is then: j ′ = a 2 sin 2 t + b 2 cos 2 t 1 ( − b cos t , − a sin t ) Now, we can express origin of S with respect to S': O ′ O = − O O ′ = − a cos t ⋅ i − b sin t ⋅ j = a 2 sin 2 t + b 2 cos 2 t a cos t ( a sin t ⋅ i ′ + b cos t ⋅ j ′ ) − a 2 sin 2 t + b 2 cos 2 t b sin t ( b cos t ⋅ i ′ − a sin t ⋅ j ′ ) = a 2 sin 2 t + b 2 cos 2 t 1 ( ( a 2 − b 2 ) sin t cos t ⋅ i ′ + a b ⋅ j ′ ) Thus, curve which is traced by the center of ellipse is: ρ = a 2 sin 2 t + b 2 cos 2 t 1 ( 2 a 2 − b 2 sin 2 t , a b ) From here, it seemed to me a very tedious task to find a closed form expression for enclosed area, so I approximated the answer and matched it with the formula which is closest to it. However, I would like to know about a nice way to derive the formula, if it exists at all. @Hosam Hajjir ?