CONICS

Definition
A conic is frequently defined in euclidean geometry as a plane section of a right circular cone OR it is
the intersection of a plane and a double-napped cone. In the formation of the four basic conics
(parabola, circle, ellipse, and hyperbola), the intersecting plane does not pass through the vertex of
the cone. This is why they are called nondegenerate conics. When the plane does pass through the
vertex, the resulting figure is a degenerate conic. The nondegenerate conics, i.e., parabola, circle,
ellipse, and hyperbola, are studied in this chapter.
Each conic section discussed here will be defined as a locus (collection) of points satisfying a
certain geometric property.

Parabolas

A Parabola is a set of all points (x, y) in a plane that are

equidistant from a fixed line, the directrix, and a fixed point, the
focus, not on the line. (See figure 1) The midpoint between the
focus and the directrix is the vertex, and the line passing through
the focus and the vertex is the axis of the parabola.
From the figure, we notice that a parabola is symmetric with
respect to its axis. Using the definition of a parabola we can derive
the following standard form of the equation of a parabola whose Figure 1
directrix is parallel to the x-axis or to the y-axis.

Standard Equation of a parabola

The standard form of the equation of a parabola with vertex at (h, k) is as follows
(x - h)2 = 4p(y – k), p≠0 Vertical axis ; directrix : y = k - p
(y - k)2 = 4p(x - h), p≠0 Horizontal axis ; directrix : x = h - p
The focus lies on the axis p units (directed distance) from the vertex. If the vertex is at the origin
(0, 0), the equation takes one of the following forms.
x2 = 4py Vertical axis
y2 = 4px Horizontal axis
 See figure 2 below

Figure 2

Example 1 : Finding the standard Equation of a Parabola

Find the standard form of the equation of the parabola with vertex (1, 0) and focus at (2, 0).
Solution
Because the axis of the parabola is horizontal, passing through (1, 0) and (2, 0), consider the
equation
(y - k)2 = 4p(x - h) where h = 1 , k = 0 , and p = 2 – 1 = 1. So, the standard form is
(y - 0)2 = 4(1)(x - 1) y2 = 4(x - 1)

Reflective property of Parabolas

Parabolas occur in a wide variety of applications. For instance, a
parabolic reflector can be formed by revolving a parabola about its
axis. The resulting surface has the property that all incoming rays
parallel to the axis are reflected through the focus of the parabola.
This is the principle behind the construction of the parabolic mirrors
used in reflecting telescopes. Conversely, the light rays emanating
from the focus of a parabolic reflector used in a flashlight are all
parallel to one another, as shown in figure 3
A line is said to be tangent to a parabola at a point on the
parabola if it intersects, but does not cross the parabola at that
Figure 3
point. As seen in figure 4, the tangent line to a parabola at a point P
makes equal angles with the following lines:
(1) The line passing through P and the focus, and
(2) The axis of the parabola
 Figure 4

Example 2 : Finding the tangent line at a point on a parabola

Find the equation of the tangent line to the parabola given by y = x2 at the point (1, 1).
Solution
For this parabola, p = ¼ and the focus is (0, ¼), as shown in the figure
5. You can find the y-intercept (0, b) of the tangent line by equating
the lengths of the two sides of the isosceles triangle shown in figure
5.

√ ( )
2
1 5
d1 = ¼ - b and d2 = ( 1−0 )2 + 1− = Note that d1 = ¼
4 4
- b rather than b – ¼. The order of subtraction for the distance is
important because the distance must be positive. Setting d 1 = d2
produces
1 5
−b=
4 4
b=−1

So the slope of the tangent line is

1−(−1)
m= =2 and the equation of the tangent line in slope-intercept form is
1−0
y=2 x−1
 Ellipses

An ellipse is a set of all points (x, y) in a plane, the sum of whose distances from two distinct points

Figure 5
(foci) is constant. [See figure 5(a)]
The line through the foci intersects the ellipse at two points called vertices. The chord joining the
vertices is the major axis, and its midpoint is the center of the ellipse. The chord perpendicular to the
major axis at the center is the minor axis. [See figure 5 (b)]
You can visualize the definition of an ellipse by using two thumbtacks placed at the foci. If the ends of
a fixed length of string are fastened to the thumbtacks and the string is drawn taut with a pencil, the
path traced by the pencil will be an ellipse.
To derive the standard form of the equation of an ellipse,
consider the ellipse in figure 6 with the following points:
center (h, k); vertices (h ± a, k); foci (h ± c, k). Note that the
center is the midpoint of the segment joining the foci.
The sum of the distances from any point on the ellipse to
the foci is constant. Using a vertex point, this constant sum is
(a + c) + (a - c) = 2a
Or simply the Length of the major axis.
Now if you let (x, y) be any point on the ellipse, the sum of the
distances between (x, y) and
Figure 6
the foci must be 2a. That is

√ [ x−( h−c ) ] +( y −k ) +√ [ x−( h+c ) ] +( y−k ) =2 a

2 2 2 2
Finally, in figure 7 you can see thatb 2=a2−c 2, which implies that the equation of the ellipse is
2 2 2 2 2 2
b ( x −h) +a ( y−k ) =a b

( x−h)2 ( y−k )2
2
+ 2
=1
a b
You would obtain a similar equation in the derivation by starting with a vertical major axis. Both
results are summarized as follows.

Standard equation of an ellipse

The standard form of the equation of an ellipse with center (h, k) and major and minor axes of
lengths 2a and 2b, respectively, where 0 ˂ b ˂ a, is

( x−h )2 ( y – k )2
2
+ 2
=1 Major axis is horizontal.
a b

( x−h)2 ( y−k )2
2
+ 2
=1 Major axis is vertical.
b a
The foci lie on the major axis, c units from the center, with c 2=a2−b 2.

If the center is at the origin (0, 0), the equation takes one of the following forms.
2 2
x y
+ =1 a˃b Major axis is horizontal.
a2 b 2
2 2
x y
2
+ 2 =1 b˂a Major axis is vertical.
b a

NB: When a = b, the resulting ellipse is a circle.

Figure 7 shows both the vertical and horizontal orientations of an ellipse.

Example 1: Finding the standard equation of an ellipse

Find the standard form of the equation of the ellipse having foci at (0, 1)
and (4, 1) and a major axis of length 6, as shown in the figure below

Figure 7

Solution
By the Midpoint formula, the center of the ellipse is (2, 1) and the distance from the center to one of
the foci is c = 2. Because 2a = 6, you know that a = 3.

Now, from c 2=a2−b 2, you have b=√ a2−c2 =√ 9−4=√5 .

Because the major axis is horizontal, the standard equation is

2 2
( x−2) ( y −1)
+ =1
32 ( √ 5)
2

Example 2: Graphing an ellipse

Graph the ellipse given by x2 + 4y2 + 6x – 8y + 9 = 0.
Solution
Begin by writing the original equation in its standard form (by completing the squares) as shown
below.
x2 + 4y2 + 6x – 8y + 9 = 0
(x2 + 6x) + 4(y2 – 2y) = -9
(x2 + 6x + 9) + 4(y2 – 2y +1) = -9 + 9 4(1)

(x + 3)2 + 4(y - 1)2 = 4

2 2
( x +3) ( y −1)
2
+ 2
=1
2 1
You now see that the center is (h, k) = (-3, 1). Because the denominator of the term is a2 = 22, the
endpoints of the major axis lie two units to the right and left of the center. Similarly, because the
denominator of the y-term is b2 = 12, the endpoints of the minor axis lie one unit up and down from
the center. The graph of this ellipse is shown in the figure against

Applications
Ellipses have many practical and aesthetic uses. For instance, machine gears, supporting arches, and
acoustic designs often involve elliptical shapes. The orbits of
satellites and planets are also ellipses.
Exercise: Application involving elliptical orbit
The moon travels about the earth in an elliptical orbit with Earth at
one focus, as shown below. The major and minor axes of the orbit
have lengths of 768 800km and 767 640km respectively. Find the
greatest and smallest distances (the apogee and perigee) from
earth’s center to the moon’s cent˂˃er.

Eccentricity
The concept of eccentricity is used to measure the ovalness of an ellipse.
c
The eccentricity e of an ellipse is given by the ratio e=
a
Note that 0 ˂ e ˂ 1 for every ellipse.
Since the foci of an ellipse are located along the major axis between the vertices and the center, it
follows that 0 ˂ c ˂ a.
Whenever the foci of an ellipse are relatively close to its center, the ellipse is nearly circular and the
ratio c/a is small [ see figure 8(a)]. On the other hand, foci distant from the center (OR close to the
vertices) produce an elongated ellipse with the ratio c/a close to 1 [see figure 8(b)]
NB: The orbit of the moon has an eccentricity of e ≈ 0.0549
The orbit of the earth has an eccentricity of e ≈ 0.0167

Figure 8

Hyperbolas

A Hyperbola is the set of all points in the plane for which the absolute values of the difference
between the distances to two fixed points, called the foci, is constant and positive. [See Figure 9(a)]
The definition of a hyperbola is similar to that of an ellipse. The difference is that for an ellipse, the
sum of the distances between the foci and a point on the ellipse is constant; whereas for a hyperbola,
the difference of the distances between the foci and a point on the hyperbola is constant.
The graph of a
hyperbola
has two

Figure 9
disconnected parts called the branches. The line through the two foci intersects the
hyperbola at two points called the vertices. The line segment connecting the vertices is the
transverse axis, and the midpoint of the transverse axis is the center of the hyperbola [See
figure 9(b)]. The development of the standard form of the equation of the hyperbola is
similar to that of an ellipse. Note that a, b, and c are related differently for hyperbolas than
for ellipses. For a hyperbola, the distance between the foci and the center is greater than
the distance between the vertices and the center.

Standard Equation of a Hyperbola

The standard form of the equation of a hyperbola with center at (h, k) is
2 2
( x−h) ( y −k )
− =1 Transverse axis is horizontal.
a2 b2
2 2
( y−k ) ( x −h)
− =1 Transverse axis is vertical.
a2 b2
The vertices are a units from the center, and the foci are c units from the center. Moreover,
c 2=a2+ b2. If the center of the hyperbola is at the origin (0, 0), the equation takes one of the
following forms.
2 2 2 2
x y y x
2
− 2 =1 Transverse axis is horizontal 2
− 2 =1 Transverse axis is vertical
a b a b
 Figure 10 shows both the horizontal and vertical orientations for a hyperbola

Figure 10

Exercise 1: Find the standard form of the equation of the hyperbola with foci (-1, 2) and
(5, 2) and vertices (0, 2) and (4, 2).

Asymptotes of a Hyperbola
Each hyperbola has two asymptotes that intersect at the center of
the hyperbola. The asymptotes pass through the corners of a
rectansgle of dimansions 2a by 2b, with its center at (h, k), as shown
in Figure 11.

Asymptotes of a Hyperbola
b
y=k ± (x−h) Asymptote for horizontal transverse axis
a
a
y=k ± (x−h) Asymptotes for vertical transverse axis
b Figure 11

The conjugate axis of a hyperbola is the line segment of length 2b joining (h, k + b) and (h, k - b) if
the transverse axis is horizontal, and the line segment of length 2b joining (h + b, k) and (h - b, k) if
the transverse axis is vertical.
Example 1: Sketching a Hyperbola
Sketch the hyperbola whose equation is 4 x2 − y 2=16

Solution

4 x2 − y 2=16
2 2
4x y 16
− =
16 16 16
 2 2
x y
2
− 2 =1 (standard form)
2 4
Because the x2-term is positive, you can conclude that the transverse axis is horizontal. So, the
vertices occur at (-2, 0) and (2, 0), the endpoints of the conjugate axis occur at (0, -4) and (0, 4),
and you can sketch the rectangle shown in the Figure A below. Finally, by drawing the
asymptotes, y = 2x and y = -2x, through the corners of this rectangle, you can complete the sketch,
as shown in Figure B below.

Figure A Figure B

Example 2: Finding the asymptotes of a Hyperbola

2 2
Sketch the hyperbola given by 4 x −3 y + 8 x +16=0 and find the equations of its
asymptotes.
Solution
2 2
4 x −3 y + 8 x +1 6=0

4 ( x 2+2 x +1 )−3 y 2=−16 +4 complete the square

2 2
y ( x +1 )
2
− 2
=1 standard form
2 (√ 3)
From this aquation you can conclude that the hyperbola has a vertical
transverse axis, is centered at (-1, 0), has vertices (-1, 2) and (-1, -2),
and has a conjugate axis with endpoints (-1 - √ 3, 0) and (-1 + √ 3, 0).
To sketch the hyperbola, draw a rectangle through these four points.
The asymptotes are the lines passing through the corners of the
rectangle, as shown in the figure in the right. Finally, using a = 2 and b
= √ 3, you can conclude that the equations of the asymptotes are
2 −2
y= (x+1) and y= (x+ 1)
√3 √3
Exercise 2: Find the standard form of the equation of the hyperbola having vertices (3, -5) and
(3, 1) and having asymptotes y=2 x−8 and y=−2 x +4 .

Eccentricity
c
As with ellipses, the eccentricity of a hyperbola is e= and because c ˃ a it follows that e ˃ 1. If
a
the eccentricity is large, the branches of the hyperbola are nearly flat, as shown in Figure 12(a). If
the eccentricity is close to 1, the branches of the hyperbola are more pointed, as shown in Figure
12(b) below.

Figure 12

Applications
Properties of hyperbolas can be used in radar and other detection systems.

General Equations of Conics

Classifying a Conic from its general Equation
The graph of A x 2+ Bxy +C y 2 + Dx+ Ey + F=0 is one of the following.
1. Circle: A=C A≠0
2. Parabola: AC = 0 A = 0 or C = 0, but not both.
3. Ellipse: AC ˃ 0 A and C have like signs.
4. Hyperbola: AC ˂ 0 A and C have unlike signs.

NB: The test above is valid if the graph is a conic. The test does not apply to equations such as
2 2
x + y =−1, whose graphs not conics.
Exercise 3: Classify the graph of each equation.
a. 4 x2 −9 x+ y−5=0
b. 4 x2 − y 2+ 8 x −6 y +4=0
c. 2 x2 + 4 y 2−4 x+ 12 y=0
d. 2 x2 +2 y 2−8 x +12 y +2=0