3.5.1 Universal Properties
Two-dimensional subsets of a 3D (double) cone surface in which the shapes are determined by the intersection of a 2D plane
General Equation
$$A⋅x^2+B⋅x⋅y+C⋅y^2+D⋅x+E⋅y+F=0$$
All conic sections will be represented without rotation until the section on rotations. In other words, $B=0$.
Discriminant & Related Properties
$$\Delta=B^2-4⋅A⋅C$$
Assuming no other values:
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If $B=0$, the function is even either vertically or horizontally (not rotated)
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If $Δ \lt 0$, the function is an ellipse
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If $Δ \lt 0$, $B=0$, and $A \lt C$, the function is an ellipse with a horizontal major axis
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If $Δ \lt 0$, $B=0$, and $A \gt C$, the function is an ellipse with a vertical major axis
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If $Δ \lt 0$, $B=0$, and $A=C$, the function is a circle
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If $Δ \lt 0$, $A=C$, $B=D=E=0$, the function is a circle centered at the origin
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If $Δ=0$, the function is a parabola
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If $C=D=F=0$, the function is a vertical parabola with the vertex at the origin
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If $A=E=F=0$, the function is a horizontal parabola with the vertex at the origin
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If $D=E=0$, the function is either an ellipse or hyperbola centered at the origin
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If $Δ \gt 0$, the function is a hyperbola
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If $Δ \gt 0$ and $A+C=0$, the function is a rectangular hyperbola, meaning the asymptotes are perpendicular
Focus, Directrix & Eccentricity
A focus/foci is a point or set of points around which a curve is guided
A Directrix is a fixed line perpendicular to the (major) axis of a function, which is determined by its focus and curvature
Eccentricity is a function's deviation from being circular, and a constant ratio given by $e=c/a$;
$$\text{eccentricity}=\frac{\text{distance from any point to the focus}}{\text{distance from any point to the directrix}}$$
Image taken from
CueMath and edited
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Two conic sections are congruent if the eccentricity of each are equal
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Circle eccentricities are always 0
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Ellipse eccentricities are always between 0 and 1
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Parabola eccentricities are always 1
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Hyperbola eccentricities are always between 1 and ∞
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Line eccentricities are always ∞
Degenerate Conics
Conics when either a 2D plane intercepts the vertex of a double cone, or the result of the general equation yields a non-function by real algebraic definition
No result |
$A⋅x^2+A⋅y^2+1=0$ |
Point |
$A⋅x^2+A⋅y^2=0$ |
Line |
$D⋅x+E⋅y+F=0$ |
Intersecting lines |
$x^2-y^2=0$ |
Parallel lines |
$x^2-1=0$ |
3.5.2 Circles
Closed curves with all points equidistant to an internal point
Circumference |
$2·\pi·r$ |
Area |
$\pi·r^2$ |
Arc length |
$\theta·r$ |
Sector area |
$\theta·r^2/2$ |
Chord length (k) |
$2·r·\sin\Big(\frac{\theta}{2}\Big)$ |
Segment area |
$\frac{r^2}{2}·\Big(\theta-2·\sin\Big(\frac{\theta}{2}\Big)·\cos\Big(\frac{\theta}{2}\Big)\Big)$ |
Conic general equation |
$A·(x^2+y^2)+D·x+E·y+F=0,A≠0$ |
Standard equation |
${(x-x_∘)}^2+{(y-y_∘)}^2=r^2$ |
Conic-standard conversions |
$x_∘=-D/(2·A)$
$y_∘=-E/(2·A)$
$r^2=(D^2+E^2-4·A·F)/(4·A^2)$ |
Focus Coordinates |
$(x_∘,y_∘)$ |
Eccentricity |
$0$ |
Directrix |
None |
Annulus
A ring formed by two concentric circles in which all features of circles apply with respect to differences involving two radii
Perimeter |
$2·\pi·(R+r)$ |
Area |
$\pi·(R^2-r^2)$ |
Sector Area |
$\theta·(R^2-r^2)/2$ |
Deductive Logic for Area
Apply the
area function of regular polygons, using the circumference as the perimeter and the radius as the apothem
$$A=(2·\pi·r)·\frac{r}{2}$$
Proof of Arc Length
The arc length is a fraction of the circumference, therefore can be found by the ratio to it and its angle
$$\frac{a}{2·\pi·r}=\frac{\theta}{2·\pi}$$
Proof of Sector Area
The sector area is a fraction of the circle area, therefore can be found by the ratio to it and its angle
$$\frac{A_S}{\pi·r^2}=\frac{\theta}{2·\pi}$$
Proof of Chord Length
Use the radius and half the chord length to form a right triangle

Use the
sine function for the angle
$$\sin\Big(\frac{\theta}{2}\Big)=\frac{k}{2·r}$$
Multiply by $2·r$
Proof of Segment Area
The segment area is the sector area minus the
triangular area between the center and chord
$$A_S=\frac{\theta·r^2}{2}-A_t$$
For the triangular area, use the chord length equation for the base and
right angle definition with respect to the radius to find the height
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$$k=2·r·\sin\Big(\frac{\theta}{2}\Big)$$
$$h=r·\cos\Big(\frac{\theta}{2}\Big)$$
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Substitute the triangular area using these values for $(k·h)/2$
$$A_S=\frac{\theta·r^2}{2}-\frac{r^2}{2}·2·\sin\Big(\frac{\theta}{2}\Big)·\cos\Big(\frac{\theta}{2}\Big)$$
Factor
$$A_S=\frac{r^2}{2}·\Big(\theta-2·\sin\Big(\frac{\theta}{2}\Big)·\cos\Big(\frac{\theta}{2}\Big)\Big)$$
Conic-Standard Conversion
Given the
conic general equation with the properties for a circle, group the $x$ terms and $y$ terms, and isolate the constant
$$A·x^2+D·x+A·y^2+E·y=-F$$
Divide by $A$
$$x^2+\frac{D}{A}·x+y^2+\frac{E}{A}·y=-\frac{F}{A}$$
Complete the square for the $x$ and $y$ terms
$$x^2+\frac{D}{A}·x+\frac{D^2}{4·A^2}+y^2+\frac{E}{A}·y+\frac{E^2}{4·A^2}=-\frac{F}{A}+\frac{D^2}{4·A^2}+\frac{E^2}{4·A^2}$$
Factor
$${\Big(x+\frac{D}{2·A}\Big)}^2+{\Big(y+\frac{E}{2·A}\Big)}^2=\frac{D^2+E^2-4·A·F}{4·A^2}$$
Equate the coefficients to the standard equation
3.5.3 Ellipses
Closed ovular curves whose points are the result of a constant sum between two internal points
Perimeter |
$4·a\lt 2·\pi·a, a\gt b$ |
Area |
$\pi·a·b$ |
Conic General Equation |
$A·x^2+C·y^2+D·x+E·y+F=0,A·C>0$ |
Standard Equation |
${(x-x_∘)}^2/a^2+{(y-y_∘)}^2/b^2=1$ |
Conic-Standard Conversions |
$A=b^2$
$C=a^2$
$D=-2·b^2·x_∘$
$E=-2·a^2·y_∘$
$F=b^2·{x_∘}^2+a^2·{y_∘}^2-a^2·b^2$
$a^2=C$
$b^2=A$
$x_∘=-D/(2·A)$
$y_∘=-E/(2·C)$ |
Orientation |
horizontal if $C>A \land a>b$
vertical if $A>C \land b>a$ |
Center Coordinates |
$(x_∘,y_∘)$ |
Foci Coordinates |
$f=\big(x_∘ \pm \sqrt{a^2-b^2}, y_∘\big),a>b$
$f=\left(x_∘,y_∘ \pm \sqrt{b^2-a^2}\right),b>a$ |
Eccentricity |
$e=f/a=\sqrt{1-b^2/a^2},a>b$
$e=f/b=\sqrt{1-a^2/b^2},b>a$ |
Directrix |
$x=\pm a/e,a>b$
$y=\pm b/e,b>a$ |
Conic-Standard Conversion
Given the standard equation
$$\frac{{(x-x_∘)}^2}{a^2}+\frac{{(y-y_∘)}^2}{b^2}=1$$
Multiply by $a^2·b^2$
$$b^2·{(x-x_∘)}^2+a^2·{(y-y_∘)}^2=a^2·b^2$$
Expand
$$b^2·x^2-2·b^2·x·x_∘+b^2·x_∘^2+a^2·y^2-2·a^2·y·y_∘+a^2·y_∘^2=a^2·b^2$$
Rearrange to appear as the conic general equation
$$b^2·x^2+a^2·y^2-2·b^2·x_∘·x-2·a^2·y_∘·y+b^2·x_∘^2+a^2·y_∘^2-a^2·b^2=0$$
Equate the coefficients to the conic equation
3.5.4 Parabolas
Open mirrored curves whose points are the same distance between a common internal point and an exterenal line.
Conic General Equations |
vertical: $A·x^2+D·x+E·y+F=0$
horizontal: $C·y^2+D·x+E·y+F=0$ |
The following are in vertical form
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Standard Equation |
$y=a·x^2+b·x+y_∘$ |
Vertex Equation |
$y=a·(x-x_∘)^2+y_∘$ |
Intercept Equation |
$y=a·(x-x_1)(x-x_2)$ |
Discriminant |
$\Delta=b^2-4·a·y_∘$
If $Δ \lt 0$, two $x$-intercepts
If $Δ=0$, one $x$-intercept
If $Δ>0$, no $x$-intercepts |
Conic-Standard Conversion |
$a=-A/E$
$b=-D/E$
$y_∘=-F/E$ |
Vertex Coordinates |
$(x_∘,y_∘)=\big(-\frac{b}{2·a},y_∘-\frac{b^2}{4·a}\big)$ |
$x$-Intercepts |
$\lbrace (x_1,0),(x_2,0) \rbrace = \big(\frac{-b \pm \sqrt{b^2-4·a·y_∘}}{2·a},0\big)$ |
Fucus Length from Vertex |
$f=\frac{1}{4·a}$ |
Focus Coordinates |
$\big(-\frac{b}{2·a},y_∘-\frac{b^2+1}{4·a}\big)$ |
Eccentricity |
$1$ |
Drectrix |
$y=-f$ |
Conic-Standard Conversion
Rearrange the conic general equation to isolate the $y$ term
$$E·y=-A·x^2-D·x-F$$
Divide by $E$
$$y=-\frac{A}{E}·x^2-\frac{D}{E}·x-\frac{F}{E}$$
Equate the coefficients to the standard equation
Standard-Vertex Conversion
Given the standard equation, isolate the $x$ terms
$$y-y_∘=a·x^2+b·x$$
Divide by $a$
$$\frac{y-y_∘}{a}=x^2+\frac{b·x}{a}$$
Complete the square
$$\frac{y-y_∘}{a}+\frac{b^2}{4·a^2}=x^2+\frac{b·x}{a}+\frac{b^2}{4·a^2}$$
Factor
$$\frac{y-y_∘}{a}+\frac{b^2}{4·a^2}={\bigg(x+\frac{b}{2·a}\bigg)}^2$$
Multiply by $a$
$$y-y_∘+\frac{b^2}{4·a}=a·{\bigg(x+\frac{b}{2·a}\bigg)}^2$$
Isolate $y$
$$y=a·{\bigg(x+\frac{b}{2·a}\bigg)}^2-\frac{b^2}{4·a}+y_∘$$
Equate the coefficients to the vertex equation
Standard-Intercept Conversion
Given the
quadratic formula with $y=0$, find the zeros of $x$
3.5.5 Hyperbolas
A mirrored set of open mirrored curvers whose points are the difference between two common internal points.
Image taken from
BYJU's and edited
Conic General Equation |
$A·x^2+C·y^2+D·x+E·y+F=0, A·C \lt 0$ |
Standard Equation |
$(x-x_∘)^2/a^2+(y-y_∘)^2/b^2=1, a^2·b^2 \lt 0$ |
Standard Equation (Real Terms Only) |
$\pm (x-x_∘)^2/a^2 \mp (y-y_∘)^2/b^2=1$ |
Conic-Standard Conversion |
$A=b^2$
$C=a^2$
$D=-2·b^2·x_∘$
$E=-2·a^2·y_∘$
$F=b^2·{x_∘}^2+a^2·{y_∘}^2-a^2·b^2$
$a^2=C$
$b^2=A$
$x_∘=-D/(2·A)$
$y_∘=-E/(2·C)$ |
Orientation |
horizontal if $C \lt 0; a \isin ℂ$
vertical if $A \lt 0; b \isin ℂ$ |
Vertices |
$(x_∘ \pm a,y_∘) \lor (x_∘,y_∘ \pm b)$ |
Focus Coordinates |
$(x_∘ \pm \sqrt{a^2+b^2},y_∘) \lor (x_∘,y_∘ \pm \sqrt{a^2+b^2})$ |
Eccentricity |
$e=f/a=\sqrt{1+b^2/a^2}$ |
Directrix |
$x=\pm a^2/f \lor y=\pm b^2/f$ |
Asymptotes |
$y= \pm b·(x-x_∘)/a+y_∘$ |
Conic-Standard Conversion
The proof is the same for
ellipses, however since by definition $A·C \lt 0$, either $a$ or $b$ must be imaginary. For the real-numbers-only representation, one of the terms is negated for $a$ and $b$ to both be positive.
3.5.6 Rotations 🔧
Conic section rotations occur exclusively when in the
general equation $B \neq 0$
$$A·x^2+B·x·y+C·y^2+D·x+E·y+F=0$$
The angle of rotation is used to identify a new axis
The original components are used to determine the new axis with respect to the angle of rotation
$$x=x'·\cos(\theta)-y'·\sin(\theta)$$
$$y=x'·\sin(\theta)+y'·\cos(\theta)$$
$$A·\big( \big)^2+B·\big( \big)·\big( \big)+C·\big( \big)^2+D·\big( \big)+E·\big( \big)+F=0$$