Proof of ellipse

Analysis: Since the ellipse is the intersection of the cylinder and the plane, this problem can be solved in space.

Make an orthographic projection of the ellipse, project it into a circle, and solve this problem in the circle.

In a circle, it can be proved by "harmonic mean × arithmetic mean = square of geometric mean"

Proof: Make an orthographic projection of an ellipse, and add ＇ to the corresponding point (my picture Forgot to add), let the angle between the straight line PQ and the plane where the circle is located be α.

∵C, D, Q∈ straight line PQ, ∴P'Q'=PQcosα, P'C'=PCcosα, P'D'=PDcosα

In circle O', Pass O' to form O'E'⊥P'D', then E' is the midpoint of C'D', ∴P'E'=? (P'C'+P'D')

Connect P'O' and intersect A'B' with F', then P'O'⊥A'B'

According to the cutting line theorem, P'A'?=P'C'×P' D'——①

∵P'A' is the tangent of ○O', ∴O'A'⊥P'A'

Rt△P'O'A' , according to the projection theorem P'F'×P'O'=P'A'?——②

In Rt△P'F'Q' and Rt△P'O'E', use triangle The function is easy to get P'Q'×P'E'=P'F'×P'O'——③

By combining the above three formulas, we can get P'Q'×P'E'= P'C'×P'D'

∵P'E' is the arithmetic mean of P'C' and P'D', P'C'×P'D' is P'C', The square of the geometric mean of P'D'

∴P'Q' is the harmonic mean of P'C' and P'D'.

That is, PQcosα is the harmonic mean of PCcosα and PDcosα

By subtracting cosα, PQ is the harmonic mean of PC and PD

Note: Harmonic mean × arithmetic mean =Proof of geometric mean?

Harmonic mean=2ab/(a+b), arithmetic mean=?(a+b)

∴Harmonic mean × Arithmetic mean=Geometric mean?

In the geometric proof of ellipse, it can often be placed in space and its orthographic projection can be used to turn it into a geometric proof of a circle. I feel that this method can be used for the geometric proof of conic sections

, but what is required is the center projection. Recommend a book "Conic Section Theory".