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An ellipse can be defined as the locus of points, the sum of whose distances to two fixed points (the foci) is constant. This is the elementary “two ...
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With given semi-major axes and semi-minor axes, we can define two foci, and therefore an ellipse, that will satisfy those axes.
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The foci have the property that the lines from the foci to a point on the curve make equal angles to the tangent. Hence, light shone from one focus r...
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The subnormal to point C on the curve is the segment from the intersection of the normal at C with the major axis, to the foot of the perpendicular dr...
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Given a point (the focus) and a line (the directrix), we examine the locus of the points whose distance from the focus is k times the distance from th...
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Where does the normal intersect the major axis?
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Given a point D in an ellipse, draw a chord through D. Find the intersection of the the tangents at the end of the chord. The locus of all such inters...
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A pair of diameters is conjugate if each is parallel to the tangents at the ends of the other. We show that the diameters of points whose parameters ...
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If we intersect the tangent at parametric location t with the axes, we observe that each intersection is independent of one of the parameters of the e...
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Take two tangents of an ellipse that are perpendicular to each other. The locus of all such points will be a circle.
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Salmon calls this the “Eccentric angle”. For an ellipse, the point at parametric location t is the point
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Take two nonparallel tangents of an ellipse, and the angle at their intersection can be calculated.
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Expressed in terms of slopes we see that the product of the slopes of the diameter and its tangent is -b^2/a^2.
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Position the end of a diameter at parametric location t. Find the angle between the chords to a point at parametric location s.
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We look at the locus of the intersections of the tangents at the end of the diameter and the focal chord through a point. This locus is a circle with ...
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We examine lines through the foci perpendicular to the tangent at a point. Take the diagonals of the trapezoid formed by the tangent, the two perpendi...
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We look at the locus of the projection of the center onto a tangent. The locus has this more suggestive equation: (X^2+Y^2)^2=(aX)^2+(bY)^2.
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The latus rectum is the chord through the focus to the ellipse perpendicular to the major axis. We look at the length of the semi latus rectum.
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The locus of the vertex of this triangle is the following 4th order curve:
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Take a chord in an ellipse and the tangent at one end. Then, take a line parallel to the chord from the center to the tangent. The locus of the inters...
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This is the directrix:
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Draw a line from the center of the ellipse to the tangent, parallel to the focal radius (line from the focus to the point of tangency). We see this h...
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The distance of any point on the conic from the focus is equal to the whole length of the ordinate at that point, produced to meet the tangent at the ...
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If a quadrilateral circumscribes a central conic, the line joining the centers of its diagonals passes through the center of the conic.
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Notice, that it is an ellipse.
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If we look at the usual pin and string ellipse construction and take the locus of the incenters of the triangle formed by the string, we see it also i...
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If we look at the usual pin and string ellipse construction and take the locus of the excenters of the triangle formed by the string, we see that two ...
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One of two parallel sides of a trapezium is given in magnitude and position, the other in magnitude. The sum of the remaining two sides is given. ...
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We create a tangent, then a line HL through the center of the ellipse parallel to the tangent. We then intersect the original tangent with the tangen...
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The intersection of the normals at the ends of a focal chord lies on a line parallel to the major axis through the midpoint of the chord.
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Maclaurin constructs a conic section by passing a triangle through 3 fixed points, while running two of its vertices along fixed lines, and taking the...
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We’ll examine a specific case. Let’s assume our triangle has vertices on the axes and on the line X+Y=10. We assume that two of the sides pass throu...
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To create a triangle with constant perimeter k, we set one leg equal to a, and then solve for the other side length. The envelop is a circle.
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It is important in this example to constrain the distance of the line from the points directly, rather than creating perpendiculars and constraining t...
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Salmon states that the radius of curvature of any conic is equal to the cube of the normal divided by the semi-parameter. We create this quantity in ...
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We generate the evolute, as the envelope of the normals. Point at parametric location t on the evolute is the center of curvature of the point at par...
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If we drop a perpendicular from the normal to the focal radius, then drop a perpendicular back onto the normal this is the center of curvature.
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We dilate an ellipse by factor k about its center. We observe that the segments of a chord of the outer ellipse cut off by the inner ellipse are equa...
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If we examine the two tangents to an inner conic which pass through a point on the outer conic, we see that each makes the same angle with the outer c...
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We take the family of ellipses with foci (a,0), (-a, 0) whose generic member we pass through the point (0,t). We look at the locus of the points of i...
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Again, we can force the tangency by reflecting one focus in the tangent line and joining the reflection to the other focus, then intersecting the resu...
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Given a triangle and a point lying on a circle concentric with the circumcircle but with twice the area, we create another triangle by reflecting the ...
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There are a family of conics that go through four points; the fifth point defines one conic. One way to determine this conic is to first make a family...
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Take a general equation, and look at the locus of midpoints of chords parallel to the y axis.
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Given a point on the conic, we create the line joining the opposite ends of the two focal chords. We examine its envelope, which is an ellipse, as w...
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Where does the normal intersect the minor axis?
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We contruct conjugate diameters. The triangle formed by the conjugate radii and the line connecting the tangecy points has constant area.
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These conics have double contact with the original conic at the line intersection points. That is they are tangential to the original conic.
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