Conic section- Shapes that are obtained by taking different plane slices through a double cone
Menaechmus (Greek mathematician) discovered conic sections around 350 BC
Write the equation for the given centre and radius
(a) centre the origin, radius 1
(1)^2 - (x-0)^2 + (y-0)^2
1= x^2 + y^2
x^2 + y^2 -1 = 0
(b) centre (3, 4), radius 3
(3)^2 = (x-3)^2 + (y-4)^2
9= (x-3)(x-3)+ (y-4)(y-4)
9 = x^2 + y^2 -6x -8y + 25
x^2 + y^2 -6x -8y +16=0
Determine the condition on the numbers f, g, and h in the equation x^2 + y^2 + fx + gy + h = 0 for the equation to pass through the origin
(c) Consider the set of points (x,y) that satisfy the equation: x^2 + y^2 -4x +6y +9 = 0
Our goal is to complete the square (note that the coefficients on x^2 and y^2 are both 1)
Re-write equations containing just the x-terms, and just the y-terms
x^2 − 4x = (x − 2)^2− 4
y^2 + 6y = (y + 3)^2 − 9
These terms can be substituted into our original equation
(x-2)^2 -4 + (y+3)^2 -9 +9 = 0
(x-2)^2 + (y+3)^2 = 4
This means that the circle has a radius of 2 and a centre of (2, -3)
Theorem 2: You can complete the square in the equation x^2 + y^2 + fx + gy + h = 0 to put it in the form (x+1/2f)^2 + (y+1/2g)^2 = 1/4 f^2 +1/4g^2 -h
This equation represents a circle with the centre (-1/2f, -1/2g) and a radius sqrt(1/4f^2 + 1/4g^2 -h)
This only occurs when the 1/4f^2 + 1/4g^2 -h is greater than zero, if this is less than zero, we can prove that the figure is not a circle
Why?
The equation is equal to r^2 (radius squared), when any number is squared, it has to equal a positive number. Also, when you go to take the square root of this side of the equation, it is impossible to get a rational answer when square-rooting a negative number.
(d) Determine the centre and radius of the equation
3x^2 + 3y^2 − 12x − 48y = 0.
(e) Determine the set of points that satisfies the following equation
x^2 + y^2 − 2x + 4y + 5 = 0;
Two intersecting circles that meet at right angles
Theorem 3: Orthogonality test
Two intersecting circles with equations x^2 + y^2 + f1x + g1 y + h1 = 0 and x^2 + y^2 + f2x + g2 y + h2 = 0, are only orthogonal if f1 f2 + g1g2 = 2(h1 + h2).
Why?
This can be tested with the pythagorem theorem, where P is a point of intersection, and A and B are points inside of each of the circles. This means that
AP^2 + BP^2 = AB^2. Using our previous formulas, we know that
This can be solved to result in the original equation
In this image, the circle intersects at distinct points P and Q
This equation represents a circle when k does not equal -1, and this means that P and Q will satisfy the equation
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These circles have distinct points of intersection, and the line containing these points can be expressed using the second equation, only when k=-1. When k is any other number, this is the equation of one of the circles
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<<The set of points (P) whose distance from a fixed point (F) is equal to their distance from a fixed line (D)<<
Standard form:
y^2=4ax
The focus and directrix can be described by
x = at^2, y = 2at (t ∈ R).
The x-axis is the axis of the parabola in standard form because the parabola is symmetric in respect to this line.
The origin is the vertex of the parabola, and it has no centre
<<e is greater than or equal to zero, but less than one<<
A set of points whose distance from a certain point is e times their distance from a fixed line
Standard form:
This equation continues to grow, however, as the figure is symmetrical.
There is a second focus (-ae, 0) and a second directrix (-a/e)
The ellipse intersects the axes at the points (±a, 0) and (0, ±b).
Parametric equations:
x = a cost, y = b sin t (t ∈ (−π, π]).
^^e is greater than 1^^
Standard form:
When: b^2 = a^2 (e^2 − 1), a > 0, e > 1.
Parametric equations: x = ct, y = c/t where t = 0.
You can describe the equation of a non-degenerate conic in terms of polar coordinates
This formula comes from the equation OP = e · PM where the distance from the origin to a point is equal to the eccentricity multiplied by the distance between that same point and the directrix.
Theorem 5: The sum of the two focal distances from a point always equal double the major axis.
Why?
The distance from a point to either foci is equal to the eccentricity times the distance from said point to the directrix. This distance is equal to the distance from the second foci to to second directrix (d’ and f’). When adding these two distances, they will always equal 2a (double the major axis)
Theorem 6: Differences of Focal Distances of Hyperbola
A slant plane (pi) that cuts one portion of a right, circular cone is an ellipse
Goals: Find the slope of the line that is tangent to the curve at the parameter, so that you can determine the equation of the tangent
Theorem 1: The slope of a tangent to a curve is y’(t)/ x’(t)
Theorem 2: Equations of tangents in standard form
Polar of (a,b) with respect to the unit circle
If (a,b) is some point on a tangent, then it can be determined that ax1 + by1 = 1 because of the equation xx1 + yy1 = 1. This equation can be re-written with points x2 and y2, which deduces the fact that points
( x1, y1) and (x2, y2) are on the same line, and this line is called the polar of (a,b) with respect to the unit circle
Theorem 3: When a point (a,b) lies outside of the circle, and the tangents from point (a,b) touch the circle at P1 and P2, then the equation of the line is ax + by = 1
Reflection law- The angle that incoming light makes with the tangent to a surface is the same as the angle that the reflected light makes with the tangent.
Reflection property of the Ellipse - light which comes from one focus of an elliptical mirror is reflected at the ellipse to pass through the second focus.
Reflection property of the hyperbola- light coming from one focus of a hyperbolic mirror is reflected at the hyperbola and makes the light appear to have come from the other focus (Internal Reflection property). Light going towards one focus is reflected towards the other focus (External Reflection Property)
Reflection Property of the Parabola- incoming light parallel to the axis is reflected at the parabola to pass through the focus. Light coming from the focus of a parabola is reflected to give a beam of light parallel to the axis of the parabola
Non-degenerate conics are the envelopes of a family of lines that are tangent to the conics
Auxiliary circle- A circle whose diameter is its major axis
Theorem 4: A perpindicular from a focus of a non-degenerate conic to a tangent meets the tangent on the auxiliary circle of the conic
Parabola:
Ellipse:
Hyperbola:
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Standard form of conics- the centre is at the origin, and the axes are parallel to the x and y axis
Matrix form of the equation: x^T Ax + J^T x + H = 0.
Look back at the old equation: Ax^2 + Bxy + Cy^2 + Fx + Gy + H = 0
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Theorem 2: A 2x2 matrix P represents a rotation of R^2 when P is orthogonal and the det of P = 1
Classify a conic (E) with the equation Ax^2 + Bxy + Cy^2 + Fx + Gy + H = 0
Write the equation in matrix form (x^T Ax + J^T x + H = 0)
Determine the orthogonal matrix P, with a determinant of 1 that diagonalizes A
Make the change of coordinate system x=Px’. This changes the equation to λ1x’^2 + λ2 y’^2 + f x’ + gy’ + h = 0
Complete the square to rewrite the equation of E into (x’’, y’’) as the equation of a conic in standard form
Use the equation x’ = P^Tx to determine the centre and axes of E in the terms of the original coordinate system.
Quadric surfaces- (quadrics) surfaces in R^3 that are the natural analogues of the conics.
Degenerate quadrics- Empty set, single point, single line, single plane, pair of planes, and a cylinder
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Idetifying matrices: In the equation Ax^2 + By^2 + Cz^2 + Fxy + Gyz + Hxz + Jx + Ky + Lz + M = 0,
If a 3x3 matrix (P) represents a rotation of R^3 about the origin when P is orthogonal and det P = 1
Classify quadric E with the equation Ax^2 + By^2 + Cz^2 + Fxy + Gyz + Hxz + Jx + Ky + Lz + M = 0,
Write the equation in matrix form (x^T Ax + J^T x + M = 0)
Determine the orthogonal matrix P, with a determinant of 1 that diagonalizes A
Make the change of coordinate system x=Px’. This changes the equation to λ1x’^2 + λ2 y’^2 + λ3 z’^2 + jx’ + ky’ + lz” +m = 0
Complete the square to rewrite the equation of E into (x’’, y’’, z’’) as the equation of a quadric in standard form
Use the equation x’ = P^Tx to determine the centre and axes of E in the terms of the original coordinate system.
Ruled Surface in R^3- a surface that can be made up from a family of straight lines
Theorem 2: A hyperboloid of one sheet contains two families of generating lines, The members of each family are disjoint, and each member of either family intersects each member of the other, with exactly one exception
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