Conic sections are the curves formed when a plane intersects the surface of a right circular cone. Except in some degenerate cases, the curves formed are the circle, the ellipse, the parabola, and the hyperbola.
Definition. A parabola is the set of all points in the plane that are the same distance from a fixed line and a fixed point not on the line. The fixed line is called the directrix, and the fixed point is called the focus.
Axis or axis of symmetry. This is the line that passes through the focus and is perpendicular to the directrix.
Vertex. This is the point at which the parabola intersects its axis. The vertex is halfway between the focus and the directrix.
Standard forms of equations of parabolas with vertex at (0, 0) andp>0are
The latus rectum of a parabola is the line segment that passes through the focus, is perpendicular to the axis of the parabola, and has endpoints on the parabola. The length of the latus rectum (focal diameter) of a parabola y2=±4px or x2=±4py is 4p.
Standard forms of equations of parabolas with vertex (h, k) and p>0 are (y−k)2=±4p(x−h) and (x−h)2=±4p(y−k). (See the table on page 667.)
Find the focus, directrix, and focal diameter (length of the latus rectum) of the parabola y2=12x.
Solution
The equation y2=12x has the form y2=4px, so the parabola opens to the right. We have 4p=12. So p=3. The focus of the parabola is (3, 0), with directrix x=−3.
The focal diameter=4p=12.
Find the standard equation of the parabola with vertex (0, 0); axis along the y-axis; and graph passing through the point (2,−3).
Solution
The axis of the parabola is the y-axis and its vertex is (0, 0); its equation has the form x2=4py or x2=−4py. The sign ± depends on whether the parabola opens up or down. Since the parabola passes through the point (2,−3), which lies in the fourth quadrant, the parabola must open down. So, its equation has the form x2=−4py.
The parabola passes through the point (2,−3), so
x2(2)2p===−4py−4p(−3)412=13
The equation of the parabola is x2=−43y.
Find the standard equation of the parabola with vertex (−2,1) and focus (−2,5).
Solution
The vertex and focus lie on the vertical line, x=−2. Also, the focus is above the vertex, so the equation has the form (x−h)2=4p(y−k). Here (h,k)=(−2,1) and p=distance between the vertex and the focus=4. The equation of the parabola is
(y−1)2=16(x+2).
7.3 The Ellipse
Definition. An ellipse is the set of all points in the plane, the sum of whose distances from two fixed points is a constant. The fixed points are called the foci of the ellipse.
Vertices. These are the two points where the line through the foci intersects the ellipse.
Major axis. This is the line segment joining the two vertices of the ellipse.
Center. This is the midpoint of the line segment joining the foci. It is also the midpoint of the major axis.
Minor axis. This is the line segment that passes through the center of the ellipse, is perpendicular to the major axis, and has endpoints on the ellipse.
Standard forms of equations of ellipses with center(0,0),a>b>0,c<a, and b2=a2−c2 are
x2a2+y2b2=1, with foci (±c,0) and vertices (±a,0),
and
x2b2+y2a2=1, with foci (0,±c) and vertices (0,±a).
Find the standard form of the ellipse with vertices (0,±5) and foci (0,±3).
Solution
Because the foci and the vertices are on the y-axis, the major axis is along the y-axis. The equation has the form x2b2+y2a2=1. We have a=5 and c=3. So b2=a2−c2=25−9=16. The equation of the ellipse is
x216+y225=1.
Find the vertices and foci of the ellipse with equation 5x2+9y2=45.
Solution
Divide both sides of 5x2+9y2=45 by 45 and simplify, to get
So a2=9 and b2=5. This is a horizontal ellipse with vertices (±a,0) and foci (±c,0). Here
c2=a2−b2=9−5=4. So c=2.
The vertices are (±3,0) and foci are (±2,0).
Find the center, vertices, and foci of the ellipse 9x2−36x+4y2+24y+36=0.
Solution
Complete the squares on x and y:
This is a vertical ellipse, with a2=9 and b2=4 and center (2,−3).
Here c2=a2−b2=9−4=5. So a=3 and c=5–√.
The vertices are (2,−3+3)=(2,0) and (2,−3−3)=(2,−6). The foci are (2,−3+5–√) and (2,−3−5–√).
7.4 The Hyperbola
Definition. A hyperbola is the set of all points in the plane, the difference of whose distances from two fixed points is a constant. The fixed points are called the foci of the hyperbola.
Vertices. These are the two points where the line through the foci intersects the hyperbola.
Transverse axis. This is the line segment joining the two vertices of the hyperbola.
Center. This is the midpoint of the line segment joining the foci. It is also the midpoint of the transverse axis.
Standard forms of equations of hyperbolas with center (0, 0),c>a, and b2=c2−a2 are
x2a2−y2b2=1, with foci (±c,0) and vertices (±a,0)
and
y2a2−x2b2=1, with foci (0,±c) and vertices (0,±a).
Conjugate axis. This is a line segment of length 2b that passes through the center of the hyperbola and is perpendicularly bisected by the transverse axis.
Asymptotes. The asymptotes of the hyperbola x2a2−y2b2=1 are the two lines y=±bax; the asymptotes of the hyperbola y2a2−x2b2=1 are the lines y=±abx.
Graphing. A procedure for graphing a hyperbola centered at (0, 0) is given on page 697.
Standard forms of equations of hyperbolas centered at (h, k) are
(x−h)2a2−(y−k)2b2=1 and (y−k)2a2−(x−h)2b2=1.
See the table on page 698. Also see page 699 for a procedure for graphing such hyperbolas.
Find the standard form of the equation of the hyperbola with vertices at (0,±2) and foci at (0,±4).
Solution
The vertices and the foci are on the y-axis, so the equation of the hyperbola has the form:
y2a2−x2b2=1.
Here a=2 and c=4. So, b2=c2−a2=16−4=12. The equation of the hyperbola is
y24−x212=1.
Find the vertices, foci, and asymptotes of the hyperbola 3x2−y2=27.
Solution
Divide both sides of 3x2−y2=27 by 27 to get the standard form:
x29−y227=1.
The coefficient of y2 is negative; the transverse axis lies on the x-axis. Here a2=9 and b2=27. So, c2=a2+b2=9+27=36. Then, a=3 and c=6. The vertices are (±a,0)=(±3,0); foci=(±c,0)=(±6,0). The asymptotes are given by x29−y227=0 or y=±3–√x.
Find the center, vertices, and foci of the hyperbola 12y2−24y−4x2−16x=52.
Solution
Complete the squares on x and y.
12(y2−2y+1)−4(x2+4x+4)12(y−1)2−4(x+2)2(y−1)24−(x+2)212===52+12−16,so481Divide by 48and simplify.
The coefficient of x2 is negative, so the transverse axis is parallel to the y-axis. Here a2=4 and b2=12. So, c2=a2+b2=4+12=16. Then, a=2 and c=4.
Center=(−2,1)
Vertices=(−2,1±a)=(−2,1±2)=(−2,3) and (−2,−1).Foci=(−2,1±c)=(−2,1±4)=(−2,5) and (−2,−3).