Geometrical properties of parabolic (traditional geometric method derivation)

Parabolic has a lot of geometrical properties, there are many articles on the derivation of these properties on the Internet, but almost exclusively is used to analyze the geometry of the method. Simultaneous equations, the derivation of the relationship between the root and the coefficient, the calculation calculation ...

However, unlike the elliptic and hyperbolic curves, which are also two curves, the geometrical properties of the circles and the parabolic lines are very "good", and many conclusions can be drawn without the coordinate method. However, compared with the perfect symmetry of the circle, the parabola is still inferior to many. The tangent of the circle is easy to describe with geometrical conditions (it is easy to use contradiction to pass the tangent of the circle perpendicular to the diameter of the cut-off point), while the tangent of the parabola is easy to describe with geometrical conditions, but the relevant conclusions are difficult to be proven by the pure geometry method. Therefore, it is necessary to use the coordinate method to prove an important conclusion when the tangent problem is involved. Nevertheless, the process of proving this paper is much more refreshing than the algebraic method with a large lump of equations.

To make a conclusion, we have to give a definition first:

defines a graphic, called a parabola , that consists of all points within a plane to a fixed point and an equal distance from a fixed line. Fixed-point is called the Focus of the parabolic line, the alignment is called the parabola, the distance of focus to the alignment is called Gio Chun distance.

Conclusion 1 parabola is an axisymmetric shape, and the perpendicular of the cross focus is an axis of symmetry.

Prove

Set the focus to \ (f\), the line is \ (l\), the axis is \ (a\), there is a point on the parabola (p\). Over \ (p\) for \ (PP ' \perp l\), perpendicular for \ (P ' \). When \ (p\) is not on \ (a\), make \ (p\) about \ (a\) the symmetry point \ (q\), make \ (P ' \) about \ (a\) the symmetry point \ (Q ' \). Connect \ (fp\), \ (fq\). by \ (a \perp l\) know \ (PP ' \parallel a \), so \ (QQ ' \parallel a \), so \ (QQ ' \perp l\). Known by symmetry (PP ' =qq ' \), \ (fp=fq\), and \ (FP=PP '), so \ (FQ=QQ ' \), so \ (q\) on the parabola, the conclusion is proof.

The vertical line defining the parabola's cross-focus is called the axis of the parabola, and the intersection of the axis and the parabolic line is called the vertex of the parabolic line.

Conclusion 2 The focus of the parabolic line is \ (f\), the vertex is \ (o\), the Gio Chun distance is \ (p\), for any point on the parabola \ (p\), \ (FP = \frac{p}{1+\cos{\angle{ofp}}}\).

Prove

Set \ (fp=\rho\), \ (\angle{ofp}=\theta\).

, when \ (\theta > 90^{\circ}\), make \ (fp\) on the axis of the projection, easy (\rho = p-\rho\cos{\theta}\). Tidy up \ (\rho = \frac{p}{1+\cos{\theta}}\), i.e. \ (FP = \frac{p}{1+\cos{\angle{ofp}}}\).

The same can be proven when \ (0^{\circ} < \theta < 90^{\circ}\), the conclusion is still true.

When \ (\theta = 90^{\circ}\), \ (pf=p \), the conclusion is still valid.

When \ (\theta = 0^{\circ}\), \ (PF = \frac{p}{2} \), the conclusion is still true.

In summary, for any point on the parabolic line \ (p\), the conclusion is established.

Inference 1 the focal distance of the parabolic line is \ (p\), the parabola focus \ (f\) of the straight line and parabola in \ (a\), \ (b\) Two points, then there is \ (\frac{1}{af}+\frac{1}{bf}=\frac{2}{p}\).

Inference 2 The vertex of the parabolic line is \ (o\), the Gio Chun distance is \ (p\), \ (\angle{ofp}=\theta\), the parabola focus \ (f\) of the straight line and parabola in the \ (a\), \ (b\) Two points, then there \ (ab=\frac{2p}{ \sin^2{\theta}}\).

Conclusion 3 the intersection of the parabola axis and the crosshair is \ (k\), the straight line and parabola of the parabolic focus \ (f\) are handed in \ (a\), \ (b\) two points, then the axes are divided equally (\angle{akb}\).

, set the guideline to \ (l\), axis \ (a\), over \ (a\) as \ (Ad\perp l\), cross \ (l\) in \ (d\), over B (Bc\perp l\), cross \ (l\) in \ (c\).

\ (\because\) \ (ad\perp l\) and \ (Bc\perp l\)

\ (\therefore\) \ (ad\parallel a\) and \ (Bc\parallel a\)

\ (\therefore\) \ (\FRAC{KD}{KC}=\FRAC{FA}{FB} \)

also \ (\because\) \ (fa=ad\) and \ (fb=bc\)

\ (\therefore\) \ (\FRAC{KD}{KC}=\FRAC{AD}{BC} \)

\ (\therefore\) \ (\triangle{kda}\sim \triangle{kcb}\)

\ (\therefore\) \ (\angle{dka} = \angle{ckb}\)

\ (\therefore\) axis split \ (\angle{akb}\)

Conclusion 4 the parabolic focus is \ (f\), the alignment is \ (l\), the intersection of the axis and the alignment is \ (k\), the line and parabola of the Cross \ (f\) in the \ (a\), \ (b\) two points, over \ (a\) as \ (Ad\perp l\), intersection (l\) In \ (d\), over B (Bc\perp l\), intersection (l\) in \ (c\), then \ (fd\) split (\angle{kfa}\), \ (fc\) split (\angle{kfb}\), \ (Fc\perp fb\).

Prove

\ (\because\) \ (fb=bc\), \ (fa=ad\)

\ (\therefore\) \ (\angle{afd}=\angle{adf}\), \ (\angle{bfc}=\angle{bcf}\)

\ (\because\) \ (kf\parallel ad\), \ (Kf\parallel bc\)

\ (\therefore\) \ (\angle{kfd}=\angle{adf}\), \ (\angle{kfc}=\angle{fcb}\)

\ (\therefore\) \ (fd\) split \ (\angle{kfa}\), \ (fc\) split (\angle{kfb}\)

\ (\therefore\) \ (Fc\perp fb\)

Conclusion 5 the parabolic focus is \ (f\), the line is \ (l\), the vertex is \ (o\), the Cross \ (f\) of the straight and parabola in the \ (a\), \ (b\) two points, over \ (a\) make \ (Ad\perp l\), intersection (l\) in \ (D \), B (Bc\perp l\), cross \ (l\) in \ (c\), then \ (a\), \ (o\), \ (c\) Three-point collinear, \ (b\), \ (o\), \ (d\) three-point collinear.

Prove

Connection \ (ac\) Cross axis (O ' \)

by \ (\triangle{ao ' F}\sim \triangle{acb}\)

\ (\frac{bc}{o ' f}= \frac{ab}{af} \)

\ (\frac{bc}{o ' f}= \frac{af+bf}{af} \)

\ (\frac{bc}{o ' f}= 1+\frac{bf}{af} \)

\ (\frac{bc}{o ' f}= 1+\frac{bc}{af} \)

\ (\frac{1}{o ' f}= \frac{1}{bc}+\frac{1}{af} \)

\ (\frac{1}{o ' f}= \frac{1}{bf}+\frac{1}{af} \)

By the conclusion 2 the inference 1 had

\ (\frac{1}{o ' f}= \frac{2}{p} \)

\ (O ' f= \frac{p}{2} \)

\ (\therefore \) \ (O ' \) coincides with \ (o\)

\ (\therefore \) \ (a\), \ (o\), \ (c\) three-point collinear

In the same vein (b\), \ (o\), \ (d\) three points collinear.

\ (\therefore \) conclusion established.

A geometrical property of a parabola tangent is proved by the coordinate method, which is used as a geometrical condition to describe the parabola tangent.

The theorem is in a planar Cartesian coordinate system \ (xoy\), with the focus being \ ((0, \frac{p}{2}) \), and the parabolic equation for \ (y=-\frac{p}{2} \) is \ (x^2=2py \).

Proof slightly.

Conclusion 6 sets the parabolic vertex to \ (o\), the tangent of the vertex is \ (l\), and the parabolic line has a point \ (p\) that is different from the vertex. Over point P is the tangent of the parabola (m\), intersection (l\) in \ (m\), \ (p\) on \ (l\) the projection is \ (P ' \), then \ (m\) is the midpoint of \ (op\).

Prove

Building a system. Set \ (P (x_0, y_0) \)

\ (y=\frac{x^2}{2p} \)

\ (\rightarrow y ' =\frac{x}{p} \)

\ (\rightarrow m:y-y_0 = \frac{x_0}{p} (x-x_0) \)

Order \ (y=0 \), get

\ (-y_0 = \frac{x_0}{p} (x-x_0) \)

Solution to

\ (x = \frac{x_0}{2} \)

\ (\therefore \) Conclusion Proof

Conclusion 7 The parabolic vertex is \ (o\), the focus is \ (f\), the line is \ (l\), the intersection of the axis and the alignment is \ (k\), the tangent of the vertex is \ (m\), and the parabola has a point that is different from the vertex \ (p\). Cross point \ (p\) as the tangent of the parabola (t\), intersection (m\) in \ (m\). (p\) (Pd\perp l\), intersection (l\) in \ (d\), then \ (f\), \ (d\), \ (m\) Three-point collinear, tangent \ (pm\) is the fd\ of \ (perpendicular bisector).

Prove

by \ (o\) is the midpoint of \ (kf\) and conclusion 6, \ (f\), \ (d\), \ (m\) three points collinear

\ (\therefore \) \ (DM=FM \)

also \ (\because \) \ (pf=pd \), \ (MP=MP \)

\ (\therefore \) \ (\triangle{pmf}\cong \TRIANGLE{PMD} \)

\ (\therefore \) \ (pm\) vertical split (fd\)

\ (\therefore \) Conclusion Proof

The light emitted from the focal point of the parabola is inferred to be emitted by a parabolic reflection, parallel to the axis of the parabolic line.

Conclusion 8 Set the parabolic focus to \ (f\), the line is \ (l\), the tangent of the vertex is \ (m\), the Cross \ (f\) of the straight and parabola in the \ (a\), \ (b\) two points, over \ (a\) as \ (Ad\perp l\), intersection (l\) in \ ( d\), B (Bc\perp l\), cross \ (l\) in \ (c\), over \ (a\), \ (b\) as the tangent of the respective parabola (t_1\), \ (t_2\), \ (T_1\cap m=g \), \ (T_2\cap m=h \), The intersection of \ (t_1\) and \ (t_2\) \ (t\) is on \ (l\), \ (tc=td\), and \ (Ta\perp tb\).

Prove

Set \ (T_1\cap l=t_1 \), \ (T_2\cap l=t_2 \).

From conclusion 4 and conclusion 7 know, \ (T_1g\perp fd\), \ (Cf\perp fd\)

\ (\therefore \) \ (T_1g\parallel cf\)

Also by conclusion 7 know, \ (g\) is the midpoint of \ (fd\)

\ (\therefore \) \ (t_1g\) is the median line of \ (\triangle{dfc}\)

\ (\therefore \) \ (t_1\) is the midpoint of \ (cd\)

In the same vein, \ (t_2\) is also the midpoint of \ (cd\)

\ (\therefore \) \ (t_1\) coincides with \ (t_2\)

\ (\therefore \) \ (t_1\) and \ (t_2\)

Set this to \ (t\).

From conclusion 4 and conclusion 7 know, \ (\angle{tgf}=\angle{gfh}=\angle{fht}=90^{\circ}\)

\ (\therefore \) \ (\angle{htg}=90^{\circ}\)

\ (\therefore \) \ (Tg\perp th\), i.e. \ (Ta\perp tb\)

To sum up, the conclusion is to be proven.

inference 1 quadrilateral \ (tgfh\) is a rectangle.

Inference 2 is tangent to a circle with a diameter of \ (ab\), with a circle with a diameter of \ (af\) and a circle with a diameter of \ (bf\) tangent to \ (m\).

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Geometrical properties of parabolic (traditional geometric method derivation)