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Conic Section 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:
Ax2
+ Bxy + Cy2 + Dx + Ey + F = 0
A ≠ 0, B ≠ 0 and C ≠ 0

The result of B2 – 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: x2/a2 + y2/b2 = 1 and x2/b2 + y2/a2 = 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: x2 + y2 = r2

•    If the result is equal to 0, then we have a parabola.

o    The parabola is defined as: y2 = 4 ∙ a ∙ x and x2 = 4 ∙ a ∙ y

•    If the result is bigger than 0, then we have a hyperbola.

o    The hyperbola is defined as: x2/a2  y2/b2 = 1 and y2/a2  x2/b2 = 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 = c2

Example

If the hyperbola passes through the focus of the ellipse x2/25 + y2/16 = 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 x2/9 – y2/16 = 1 and foci is (5,0)

The equation of the hyperbola is x2/9 – y2/25 = 1 and foci is (5,0)

The equation of the hyperbola is x2/25 – y2/16 = 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:

x2/a2 + y2/b2= 1

or

x2/b2 + y2/a2 = 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 c2 = a2 - b2.

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 + b2/a2)

Summary of Basic Properties

 Circle Ellipse Parabola Hyperbola Standard Cartesian Equation : x2 + y2 = r2 x2/a2 + y2/b2= 1 y2 = 4ax x2/a2 -  y2/b2= 1 Eccentricity (e): 0 0 < e <1 1 1 < e Relation between a,b and e b = a b2 = a2(1-e2) b2 = a2(e2-1) Parametric Representation x = a · cosθ y = b · sinθ x = at2 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

Ax2 + Bxy + Cy2 + Dx + Ey + F = 0
A ≠ 0, B ≠ 0 and C ≠ 0

The actual type of conic can be found from the sign of B2 - 4AC

 B2 - 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

Ax2 + 2Bxy + Cy2 + Dx + Ey + F = 0

Here the type of conic must be found from the sign of B2 - AC

 B2 - 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

x2/a2 +  y2/b2= 1

where a and b would give the lengths of the semi-major and semi-minor 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: 2b2 / 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 x2/a2 – y2/b2 = 1 it has:

foci (±c, 0), where c2 = a2 + b2

vertices (±a, 0)

and asymptotes y = ±(b/a)x.

For hyperbola y2/a2 – x2/b2 = 1 it has:

foci (0, ±c), where c2 = a2 + b2

vertices (0, ±a),

and asymptotes y = ±(a/b)x

In this example we will use the second formula namely y2/a2 – x2/b2 = 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 c2 = a2 + b= 5/4 and the foci are (0, ±√5/2). Inserting these values will give us the final equation of the hyperbola, namely y2 - 4x2 = 1.

Example

Find the focus and directrix for y2 = -2x.

From the given equation of the parabola, we have x = - y2/2, 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.

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