Conic sections

Conic sections are curves created by the intersection of a plane and a cone. There are six types of conic section: the circle, ellipse, hyperbola, parabola, a pair of intersecting straight lines and a single point.

All conics (as they are known) have at least two foci, although the two may coincide or one may be at infinity. They may also be defined as the locus of a point moving between a point and a line, a directrix, such that the ratio between the distances is constant. This ratio is known as "e", or eccentricity.

Ellipses


An ellipse is a locus where the sum of the distances to two foci is kept constant. This sum is also equivalent to the major axis of the ellipse - the major axis being longer of the two lines of symmetry of the ellipse, running through both foci. The eccentricity of an ellipse is less than one.

In Cartesian coordinates, if an ellipse is centered at (h,k), the equation for the ellipse is


 * $$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$$   (equation 1)

The lengths of the major and minor axes (also known as the conjugate and transverse) are "a" and "b" respectively.

Exercise 1. Derive equation 1. (hint)

A circle circumscribed about the ellipse, touching at the two end points of the major axis, is known as the auxiliary circle. The latus rectum of an ellipse passes through the foci and is perpendicular to the major axis.

From a point P($$x_1$$, $$y_1$$) tangents will have the equation:


 * $$\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1$$

And normals:


 * $$\frac{xa^2}{x_1} - \frac{yb^2}{y_1} = a^2 - b^2$$

Likewise for the parametric coordinates of P, (a $$\cos \alpha$$, b $$\sin \alpha$$),


 * $$\frac{x\cos\alpha}{a} + \frac{y\sin\alpha}{b} = 1$$

Properties of Ellipses
S and S' are typically regarded as the two foci of the ellipse. Where $$a > b$$, these become (ae, 0) and (-ae, 0) respectively. Where $$a < b$$ these become (0, be) and (0, -be) respectively.

A point P on the ellipse will move about these two foci ut $$|PS + PS'| = 2a$$

Where a > b, which is to say the Ellipse will have a major-axes parallel to the x-axis:

$$b^2 = a^2(1-e^2)$$

The directrix will be: $$x = \pm \frac{a}{e}$$

Circles
A circle is a special type of the ellipse where the foci are the same point.

Hence, the equation becomes:

$$x^2+y^2 = r^2$$

Where 'r' represents the radius. And the circle is centered at the origin (0,0)

Hyperbolas
A special case where the eccentricity of the conic shape is greater than one.

Centered at the origin, Hyperbolas have the general equation:


 * $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$

A point P on will move about the two foci ut $$|PS - PS'| = 2a$$

The equations for the tangent and normal to the hyperbola closely resemble that of the ellipse.

From a point P($$x_1$$, $$y_1$$) tangents will have the equation:


 * $$\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$$

And normals:


 * $$\frac{xa^2}{x_1} + \frac{yb^2}{y_1} = a^2 + b^2$$

The directrixes (singular directrix) and foci of hyperbolas are the same as those of ellipses, namely directrixes of $$ x = \pm \frac{a}{e} $$  and foci of   $$ ( \pm ae, 0) $$

The asymptotes of a hyperbola lie at $$ y = \pm \frac {b}{a}x $$

Rectangular Hyperbolas
Rectangular Hyperbolas are special cases of hyperbolas where the asymptotes are perpendicular. These have the general equation:

$$ xy = c $$

Conic sections generally
Within the two dimensional space of Cartesian Coordinate Geometry a conic section may be located anywhere and have any orientation.

This section examines the parabola, ellipse and hyperbola, showing how to calculate the equation of the conic section, and also how to calculate the foci and directrices given the equation.

Deriving the equation
The curve is defined as a point whose distance to the focus and distance to a line, the directrix, have a fixed ratio, eccentricity $$e.$$ Distance from focus to directrix must be non-zero.

Let the point have coordinates $$(x,y).$$

Let the focus have coordinates $$(p,q).$$

Let the directrix have equation $$ax + by + c = 0$$ where $$a^2 + b^2 = 1.$$

Then $$e = \frac {\text{distance to focus}}{\text{distance to directrix}}$$ $$= \frac{\sqrt{(x-p)^2 + (y-q)^2}}{ax + by + c}$$

$$e(ax + by + c) = \sqrt{(x-p)^2 + (y-q)^2}$$

Square both sides: $$(ax + by + c)(ax + by + c)e^2 = (x-p)^2 + (y-q)^2$$

Rearrange: $$(x-p)^2 + (y-q)^2 - (ax + by + c)(ax + by + c)e^2 = 0\ \dots\ (1).$$

Expand $$(1),$$ simplify, gather like terms and result is:

$$Ax^2 + By^2 + Cxy + Dx + Ey + F = 0$$ where:

$$X = e^2$$

$$A = Xa^2 - 1$$

$$B = Xb^2 - 1$$

$$C = 2Xab$$

$$D = 2p + 2Xac$$

$$E = 2q + 2Xbc$$

$$F = Xc^2 - p^2 - q^2$$

Note that values $$A,B,C,D,E,F$$ depend on:


 * $$e$$ non-zero. This method is not suitable for circle where $$e = 0.$$


 * $$e^2.$$ Sign of $$e \pm$$ is not significant.


 * $$(ax + by + c)^2.\ ((-a)x + (-b)y + (-c))^2$$ or $$((-1)(ax + by + c))^2$$ and $$(ax + by + c)^2$$ produce same result.

For example, directrix $$0.6x - 0.8y + 3 = 0$$ and directrix $$-0.6x + 0.8y - 3 = 0$$ produce same result.

Parabola
Every parabola has eccentricity $$e = 1.$$

Simple quadratic function:

Let focus be point $$(0,1).$$

Let directrix have equation: $$y = -1$$ or $$(0)x + (1)y + 1 = 0.$$

As conic section curve has equation: $$(-1)x^2 + (0)y^2 + (0)xy + (0)x + (4)y + (0) = 0$$

Curve is quadratic function: $$4y = x^2$$ or $$y = \frac{x^2}{4}$$

For a quick check select some random points on the curve:

Gallery
Curve in Figure 1 below has:


 * Directrix: $$y = -23$$


 * Focus: $$(7,-21)$$


 * Equation: $$(-1)x^2 + (0)y^2 + (0)xy + (14)x + (4)y + (39) = 0$$ or $$y = \frac{x^2 - 14x - 39}{4}$$

Curve in Figure 2 below has:


 * Directrix: $$x = 12$$


 * Focus: $$(10,-7)$$


 * Equation: $$(0)x^2 + (-1)y^2 + (0)xy + (-4)x + (-14)y + (-5) = 0$$ or $$x = \frac{-(y^2 + 14y + 5)}{4}$$

Curve in Figure 3 below has:


 * Directrix: $$(0.6)x - (0.8)y + (2.0) = 0$$


 * Focus: $$(6.6, 6.2)$$


 * Equation: $$-(0.64)x^2 - (0.36)y^2 - (0.96)xy + (15.6)x + (9.2)y - (78) = 0$$

Ellipse
Every ellipse has eccentricity $$1 > e > 0.$$

A simple ellipse:

Let focus be point $$(p,q)$$ where $$p,q = -1,0$$

Let directrix have equation: $$(1)x + (0)y + 16 = 0$$ or $$x = -16.$$

Let eccentricity $$e = 0.25$$

Ellipse has center at origin and equation: $$(0.9375)x^2 + (1)y^2 = (15).$$

Some basic checking:

The effect of eccentricity.

All ellipses in diagram have:


 * Focus at point $$(-1,0)$$


 * Directrix with equation $$x = -16.$$

Five ellipses are shown with eccentricities varying from $$0.1$$ to $$0.9.$$

Gallery
Curve in Figure 1 below has:


 * Directrix: $$x = -10$$


 * Focus: $$(3,0)$$


 * Eccentricity: $$e = 0.5$$


 * Equation: $$(-0.75)x^2 + (-1)y^2 + (0)xy + (11)x + (0)y + (16) = 0$$

Curve in Figure 2 below has:


 * Directrix: $$y = -12$$


 * Focus: $$(7,-4)$$


 * Eccentricity: $$e = 0.7$$


 * Equation: $$(-1)x^2 + (-0.51)y^2 + (0)xy + (14)x + (3.76)y + (5.56) = 0$$

Curve in Figure 3 below has:


 * Directrix: $$(0.6)x - (0.8)y + (2.0) = 0$$


 * Focus: $$(8,5)$$


 * Eccentricity: $$e = 0.9$$


 * Equation: $$(-0.7084)x^2 + (-0.4816)y^2 + (-0.7776)xy + (17.944)x + (7.408)y + (-85.76) = 0$$

Hyperbola
Every hyperbola has eccentricity $$e > 1.$$

A simple hyperbola:

Let focus be point $$(p,q)$$ where $$p,q = 0,-9$$

Let directrix have equation: $$(0)x + (1)y + 4 = 0$$ or $$y = -4.$$

Let eccentricity $$e = 1.5$$

Hyperbola has center at origin and equation: $$(1.25)y^2 - x^2 = 45.$$

Some basic checking:

The effect of eccentricity.

All hyperbolas in diagram have:


 * Focus at point $$(0,-9)$$


 * Directrix with equation $$y = -4.$$

Six hyperbolas are shown with eccentricities varying from $$1.5$$ to $$20.$$

Gallery
Curve in Figure 1 below has:


 * Directrix: $$y = 6$$


 * Focus: $$(0,1)$$


 * Eccentricity: $$e = 1.5$$


 * Equation: $$(-1)x^2 + (1.25)y^2 + (0)xy + (0)x + (-25)y + (80) = 0$$

Curve in Figure 2 below has:


 * Directrix: $$x = 1$$


 * Focus: $$(-5,6)$$


 * Eccentricity: $$e = 2.5$$


 * Equation: $$(5.25)x^2 + (-1)y^2 + (0)xy + (-22.5)x + (12)y + (-54.75) = 0$$

Curve in Figure 3 below has:


 * Directrix: $$(0.8)x + (0.6)y + (2.0) = 0$$


 * Focus: $$(-28,12)$$


 * Eccentricity: $$e = 1.2$$


 * Equation: $$(-0.0784)x^2 + (-0.4816)y^2 + (1.3824)xy + (-51.392)x + (27.456)y + (-922.24) = 0$$

Reversing the process
The expression "reversing the process" means calculating the values of $$e,$$ focus and directrix when given the equation of the conic section, the familiar values $$A,B,C,D,E,F.$$

Consider the equation of a simple ellipse: $$0.9375 x^2 + y^2 = 15.$$ This is a conic section where $$A,B,C,D,E,F = -0.9375, -1, 0, 0, 0, 15.$$

This ellipse may be expressed as $$15 x^2 + 16 y^2 = 240,$$ a format more appealing to the eye than numbers containing fractions or decimals.

However, when this ellipse is expressed as $$-0.9375x^2 - y^2 + 15 = 0,$$ this format is the ellipse expressed in "standard form," a notation that greatly simplifies the calculation of $$a,b,c,e,p,q.$$

Modify the equations for $$A,B,C$$ slightly:

$$KA = Xaa - 1$$ or $$Xaa = KA + 1\ \dots\ (1)$$

$$KB = Xbb - 1$$ or $$Xbb = KB + 1\ \dots\ (2)$$

$$KC = 2Xab\ \dots\ (3)$$

$$(3)\ \text{squared:}\ KKCC = 4XaaXbb\ \dots\ (4)$$

In $$(4)$$ substitute for $$Xaa, Xbb:$$ $$C^2 K^2 = 4(KA+1)(KB+1)\ \dots\ (5)$$

$$(5)$$ is a quadratic equation in $$K:\ (a\_)K^2 + (b\_) K + (c\_) = 0$$ where:

$$a\_ = 4AB - C^2$$

$$b\_ = 4(A+B)$$

$$c\_ = 4$$

Because $$(5)$$ is a quadratic equation, the solution of $$(5)$$ may contain an unwanted value of $$K$$ that will be eliminated later.

From $$(1)$$ and $$(2):$$

$$Xaa + Xbb = KA + KB + 2$$

$$X(aa + bb) = KA + KB + 2$$

Because $$aa + bb = 1,\ X = KA + KB + 2$$

More calculations
The values $$D,E,F:$$

$$D = 2p + 2Xac;\ 2p = (D - 2Xac)$$

$$E = 2q + 2Xbc;\ 2q = (E - 2Xbc)$$

$$F = Xcc - pp - qq\ \dots\ (10)$$

$$(10)*4:\ 4F = 4Xcc - 4pp - 4qq\ \dots\ (11)$$

In $$(11)$$ replace $$4pp, 4qq:\ 4F = 4Xcc - (D - 2Xac)(D - 2Xac) - (E - 2Xbc)(E - 2Xbc)\ \dots\ (12)$$

Expand $$(12),$$ simplify, gather like terms and result is quadratic function in $$c:$$

$$(a\_)c^2 + (b\_)c + (c\_) = 0\ \dots\ (14)$$ where:

$$a\_ = 4X(1 - Xaa - Xbb)$$

$$aa + bb = 1,$$ Therefore:

$$a\_ = 4X(1 - X)$$

$$b\_ = 4X(Da + Eb)$$

$$c\_ = -(D^2 + E^2 + 4F)$$

For parabola, there is one value of $$c$$ because there is one directrix.

For ellipse and hyperbola, there are two values of $$c$$ because there are two directrices.

Parabola
Given equation of conic section: $$16x^2 + 9y^2 - 24xy + 410x - 420y + 3175 = 0.$$

Calculate $$\text{eccentricity, focus, directrix.}$$ interpreted as:

Directrix: $$0.6x + 0.8y + 3 = 0$$

Eccentricity: $$e = 1$$

Focus: $$p,q = -10,6$$

Because eccentricity is $$1,$$ curve is parabola.

Because curve is parabola, there is one directrix and one focus.

For more insight into the method of calculation and also to check the calculation:

Ellipse
Given equation of conic section: $$481x^2 + 369y^2 - 384xy + 5190x + 5670y + 7650 = 0.$$

Calculate $$\text{eccentricity, foci, directrices.}$$ interpreted as:

Directrix 1: $$0.6x + 0.8y - 3 = 0$$

Eccentricity: $$e = 0.8$$

Focus 1: $$p,q = -3, -3$$

Directrix 2: $$0.6x + 0.8y + 37 = 0$$

Eccentricity: $$e = 0.8$$

Focus 2: $$p,q = -18.36, -23.48$$

Because eccentricity is $$0.8,$$ curve is ellipse.

Because curve is ellipse, there are two directrices and two foci.

For more insight into the method of calculation and also to check the calculation:

Hyperbola
Given equation of conic section: $$7x^2 + 0y^2 - 24xy + 90x + 216y - 81 = 0.$$

Calculate $$\text{eccentricity, foci, directrices.}$$ interpreted as:

Directrix 1: $$0.6x + 0.8y - 3 = 0$$

Eccentricity: $$e = 1.25$$

Focus 1: $$p,q = 0, -3$$

Directrix 2: $$0.6x + 0.8y - 22.2 = 0$$

Eccentricity: $$e = 1.25$$

Focus 2: $$p,q = 18, 21$$

Because eccentricity is $$1.25,$$ curve is hyperbola.

Because curve is hyperbola, there are two directrices and two foci.

For more insight into the method of calculation and also to check the calculation:

Slope of curve
Given equation of conic section: $$Ax^2 + By^2 + Cxy + Dx + Ey + F = 0,$$

differentiate both sides with respect to $$x.$$

$$2Ax + B(2yy') + C(xy' + y) + D + Ey' = 0$$

$$2Ax + 2Byy' + Cxy' + Cy + D + Ey' = 0$$

$$2Byy' + Cxy' + Ey' + 2Ax + Cy + D = 0$$

$$y'(2By + Cx + E) = -(2Ax + Cy + D)$$

$$y' = \frac{-(2Ax + Cy + D)}{Cx + 2By + E}$$

For slope horizontal: $$2Ax + Cy + D = 0.$$

For slope vertical: $$Cx + 2By + E = 0.$$

For given slope $$m = \frac{-(2Ax + Cy + D)}{Cx + 2By + E}$$

$$m(Cx + 2By + E) = -2Ax - Cy - D$$

$$mCx + 2Ax + m2By + Cy + mE + D = 0$$

$$(mC + 2A)x + (m2B + C)y + (mE + D) = 0.$$

y = f(x)
Consider conic section: $$(-1)x^2 + (0)y^2 + (0)xy + (14)x + (4)y + (39) = 0.$$

This is quadratic function: $$y = \frac{x^2 - 14x - 39}{4}$$

Slope of this curve: $$m = y' = \frac{2x - 14}{4}$$

Produce values for slope horizontal, slope vertical and slope $$5:$$ $$$$$$$$$$$$$$$$$$$$ Check results:

x = f(y)
Consider conic section: $$(0)x^2 + (-1)y^2 + (0)xy + (-4)x + (-14)y + (-5) = 0.$$

This is quadratic function: $$x = \frac{-(y^2 + 14y + 5)}{4}$$

Slope of this curve: $$\frac{dx}{dy} = \frac{-2y - 14}{4}$$

$$m = y' = \frac{dy}{dx} = \frac{-4}{2y + 14}$$

Produce values for slope horizontal, slope vertical and slope $$0.5:$$ $$$$$$$$$$$$$$$$$$$$ Check results:

Parabola
Consider conic section: $$(9)x^2 + (16)y^2 + (-24)xy + (104)x + (28)y + (-144) = 0.$$

This curve is a parabola.

Produce values for slope horizontal, slope vertical and slope $$2:$$ Because all 3 lines are parallel to axis, all 3 lines have slope $$\frac{3}{4}.$$

Produce values for slope horizontal, slope vertical and slope $$0.75:$$ Axis has slope $$0.75$$ and curve is never parallel to axis.

Ellipse
Consider conic section: $$(1771)x^2 + (1204)y^2 + (1944)xy + (-44860)x + (-18520)y + (214400) = 0.$$

This curve is an ellipse.

Produce values for slope horizontal, slope vertical and slope $$-1:$$ Because curve is closed loop, slope of curve may be any value including $$\frac{1}{0}.$$

If slope of curve is given as $$\frac{1}{0},$$ it means that curve is vertical at that point and tangent to curve has equation $$x = k.$$

For any given slope there are always 2 points on opposite sides of curve where tangent to curve at each of those points has the given slope.

Hyperbola
Consider conic section: $$(-351)x^2 + (176)y^2 + (-336)xy + (4182)x + (-3824)y + (-16231) = 0.$$

This curve is a hyperbola.

Produce values for slope horizontal, slope vertical and slope $$2:$$

Asymptotes of Hyperbola
Let the conic section have the familiar equation: $$Ax^2 + By^2 + Cxy + Dx + Ey + F = 0.$$

Let a line have equation: $$y = f(x) = mx + c.$$

Let $$f(x)$$ intersect the conic section. For $$y$$ in equation of conic section substitute $$(mx + c).$$

$$y = g(x) = Ax^2 + B(mx+c)^2 + Cx(mx+c) + Dx + E(mx+c) + F = 0.$$

Expand $$g(x),$$ simplify, gather like terms and result is quadratic function in $$x:$$

$$h(x) = (a$$_$$)x^2 + (b$$_$$)x + (c$$_$$) = 0$$ where:

$$a$$_ $$ = A + Bm^2 + Cm$$

$$b$$_ $$ = 2Bcm + cC + D + Em$$

$$c$$_ $$ = Ec + F + Bc^2$$

If line $$f(x)$$ is an asymptote, then $$a$$_ $$ = b$$_ $$ = 0,$$ in which case:

$$j(m) =\ ($$_$$a)m^2 + ($$_$$b)m + ($$_$$c) = 0$$ where:

_$$a = B;\ $$_$$b = C;\ $$_$$c = A$$

and $$m_1,m_2$$ (roots of $$j(m)$$) are the slopes of the 2 asymptotes.

Latera recta et cetera
"Latus rectum" is a Latin expression meaning "straight side." According to Google, the Latin plural of "latus rectum" is "latera recta," but English allows "latus rectums" or possibly "lati rectums." The title of this section is poetry to the eyes and music to the ears of a Latin student and this author hopes that the gentle reader will permit such poetic licence in a mathematical topic.

The translation of the title is "Latus rectums and other things." This section describes the calculation of interesting items associated with the ellipse: latus rectums, major axis, minor axis, focal chords, directrices and various points on these lines.

When given the equation of an ellipse, the first thing is to calculate eccentricity, foci and directrices as shown above. Then verify that the curve is in fact an ellipse.

From these values everything about the ellipse may be calculated. For example:

Consider conic section: $$1771x^2 + 1204y^2 + 1944xy -44860x - 18520y + 214400 = 0.$$

This curve is ellipse with random orientation.

Major axis
Techniques similar to above can be used to calculate points $$I2, ID2.$$

Latus rectums
Techniques similar to above can be used to calculate points $$R, S.$$

Checking
All interesting points have been calculated without using equations of any of the relevant lines.

However, equations of relevant lines are very useful for testing, for example:


 * Check that points $$ID2, I2, F2, M, F1, I1, ID1$$ are on axis.


 * Check that points $$R, F2, S$$ are on latus rectum through $$F2.$$


 * Check that points $$Q, M, T$$ are on minor axis through $$M.$$


 * Check that points $$P, F1, U$$ are on latus rectum through $$F1.$$

Test below checks that 8 points $$I1, I2, P, Q, R, S, T, U$$ are on ellipse and satisfy eccentricity $$e = 0.9.$$

Note the differences between "raw" values of $$e_1$$ and "clean" values of $$e_2.$$

Other resources

 * Should the contents of this Wikiversity page be merged into the related Wikibooks modules such as Conic Sections/Ellipse?