By intersecting a cone with a plane, a curve is obtained and is named conic section, which is the red area shown above.
The expression for a conic section in the Cartesian coordinate system is defined as:
Ax^{2} + Bxy + Cy^{2} + Dx + Ey + F = 0
A ≠ 0, B ≠ 0 and C ≠ 0
The result of B^{2} – 4AC determines the type of the conic section obtained:
• If the result is smaller than 0, then we have an ellipse, unless the conic is degenerate.
o The ellipse is defined as: x^{2}/a^{2} + y^{2}/b^{2} = 1 and x^{2}/b^{2} + y^{2}/a^{2} = 1
• If the result is smaller than 0 and A=C, B=0 then we have a perfect circle.
o The circle is defined as: x^{2} + y^{2} = r^{2}
• If the result is equal to 0, then we have a parabola.
o The parabola is defined as: y^{2} = 4 ∙ a ∙ x and x^{2} = 4 ∙ a ∙ y
• If the result is bigger than 0, then we have a hyperbola.
o The hyperbola is defined as: x^{2}/a^{2} – y^{2}/b^{2} = 1 and y^{2}/a^{2} – x^{2}/b^{2} = 1
• If the result is bigger than 0 and A+C = 0, then we have a rectangular hyperbola.
o The rectangular hyperbola is defined as: x ∙ y = c^{2}^{
}
Example
If the hyperbola passes through the focus of the ellipse x^{2}/25 + y^{2}/16 = 1 and its transverse and conjugate axes coincide with the major and minor axes of the ellipse, and the product of the eccentricities is 1, then:
The equation of the hyperbola is x^{2}/9 – y^{2}/16 = 1 and foci is (5,0)
The equation of the hyperbola is x^{2}/9 – y^{2}/25 = 1 and foci is (5,0)
The equation of the hyperbola is x^{2}/25 – y^{2}/16 = 1 and foci is (5,0)
Focus of the hyperbola is (3,0)
NOTE:
An ellipse with centre at the origin (0,0) is the graph of:
x^{2}/a^{2} + y^{2}/b^{2}= 1
or
x^{2}/b^{2} + y^{2}/a^{2} = 1,
where a > b > 0.
The length of the major axis is 2a, and the length of the minor axis is 2b. The two foci (foci is the plural of focus) are at (± c, 0) or at (0, ± c), where c^{2} = a^{2}  b^{2}.
The eccentricy of an ellipse is a measure of how nearly circular the ellipse. Eccentricity is found by the following formula:
Eccentricity = c/a,
where c is the distance from the centre to the focus of the ellipse and a is the distance from the centre to a vertex.
The eccentricity of the hyperbola e is given by the relation e = sqrt (1 + b^{2}/a^{2})
Summary of Basic Properties

Circle

Ellipse

Parabola

Hyperbola

Standard Cartesian Equation :

x^{2} + y^{2} = r^{2}

x^{2}/a^{2} + y^{2}/b^{2}= 1

y^{2} = 4ax

x^{2}/a^{2}  y^{2}/b^{2}= 1

Eccentricity (e):

0

0 < e <1

1

1 < e

Relation between a,b and e

b = a

b^{2} = a^{2}(1e^{2})


b^{2} = a^{2}(e^{2}1)

Parametric Representation


x = a · cosθ
y = b · sinθ

x = at^{2}
y = 2at

x = a · secθ
y = b · tanθ
or
x = ½ * a(t + 1/t)
y = ½ * b(t  1/t)

Definition: It is the locus of all points which meet the condition...

distance to the origin is constant

sum of distances to each focus is constant

distance to focus = distance to directrix

difference between distances to each foci is constant

It might tidy the logic up to consider a circle to be a special case of an ellipse. Then there are two 'main' classes
an ellipse, with e < 1
a hyperbola, with e > 1
and a 'critical' class  the parabola with e = 1.
The General Equation for a Conic is
Ax^{2} + Bxy + Cy^{2} + Dx + Ey + F = 0
A ≠ 0, B ≠ 0 and C ≠ 0
The actual type of conic can be found from the sign of B^{2}  4AC
B^{2}  4AC

Then the curve is

< 0

Ellipse, circle, point or no curve.

= 0

Parabola, 2 parallel lines, 1 line or no curve.

> 0

Hyperbola or 2 intersecting lines.

Note: the above notation brings a close analogy with the formulas of quadratic equations. Sometimes, however, the formula is stated slightly differently
Ax^{2} + 2Bxy + Cy^{2} + Dx + Ey + F = 0
Here the type of conic must be found from the sign of B^{2}  AC
B^{2}  AC

Then the curve is

< 0

Ellipse, circle, point or no curve.

= 0

Parabola, 2 parallel lines, 1 line or no curve.

> 0

Hyperbola or 2 intersecting lines.

Polar Form
For an origin at a focus, the general polar form (apart from a circle) is
1 + e · cosθ = L / r
where L is the semi latus rectum.
Ellipse
The Cartesian equation of an ellipse is
x^{2}/a^{2} + y^{2}/b^{2}= 1
where a and b would give the lengths of the semimajor and semiminor axes. In its general form, with the origin at the centre of coordinates we have:
the foci are at (±ae, 0), where e is the eccentricity of the ellipse

the directrix are at x = ±a/e, where e is the eccentricity of the ellipse

the major axis of length: 2a

the minor axis is of length: 2b

the semi latus rectum is of length: 2b^{2} / a

Example
Find the foci and the equation of the hyperbola with the vertices (0,±1) and asymptote y = 2x
Recall the formulas for hyperbola mentioned in the beginning:
For hyperbola x^{2}/a^{2} – y^{2}/b^{2} = 1 it has:
foci (±c, 0), where c^{2} = a^{2} + b^{2}
vertices (±a, 0)
and asymptotes y = ±(b/a)x.
For hyperbola y^{2}/a^{2} – x^{2}/b^{2} = 1 it has:
foci (0, ±c), where c^{2} = a^{2} + b^{2}
vertices (0, ±a),
and asymptotes y = ±(a/b)x
In this example we will use the second formula namely y^{2}/a^{2} – x^{2}/b^{2} = 1 to determine its equation.
We can see that a=1 and a/b = 2 according to the values given in the example. Thus b=a/2 = 1/2 and c^{2} = a^{2} + b^{2 }= 5/4 and the foci are (0, ±√5/2). Inserting these values will give us the final equation of the hyperbola, namely y^{2}  4x^{2} = 1.
Example
Find the focus and directrix for y^{2} = 2x.
From the given equation of the parabola, we have x =  y^{2}/2, which is obtained by isolating x from the equation. The minus sign flips the parabola to the left, so we have: 1/4p = 1/2 → p = 1/2. Therefore, focus and directrix are (1/2, 0) and x = 1/2.
Below is a sketch of the parabola.