MA001 Study Guide

Unit 10: Introduction to Conic Sections

10a. write the equation of ellipses, hyperbolas, and parabolas in standard form

Compare the equations of ellipses, hyperbolas, and parabolas in standard form.
Compare how the equations of ellipses, hyperbolas, and parabolas are modified depending on their orientation.
Identify the critical components of ellipses, hyperbolas, and parabolas, such as axes of symmetry, vertices, center, directrices, or asymptotes, as appropriate. How do these critical components help to write an equation for the ellipse, hyperbola, or parabola?

An ellipse is the set of all points $(x, y)$ in the plane such that the sum of their distances from two fixed points, called foci, is a constant. The standard forms of an ellipse depend on whether the center is at the origin or off the origin and on whether the ellipse is oriented horizontally or vertically, corresponding to whether the ellipse is longer in the direction of the $x$ -axis or in the direction of the $y$ -axis, respectively. The critical components of the ellipse help identify the appropriate standard form of the equation as indicated in the table below.

ELLIPSES	Longer horizontally with major axis on x-axis	Longer horizontally with major axis parallel to x-axis	Longer vertically with major axis on y-axis	Longer vertically with major axis parallel to y-axis
Equation	$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ $a>b$	$\frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}$ $a>b$	$\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$ $a>b$	$\begin{aligned}&\frac{(x-h)^{2}}{b^{2}} \\&+\frac{(y-k)^{2}}{a^{2}}=1 \\&a>b\end{aligned}$
Center	(0, 0)	(h, k)	(0, 0)	(h, k)
Length of major axis	2a	2a	2a	2a
Length of minor axis	2b	2b	2b	2b
Coordinates of vertices	(-a, 0) and (a, 0)	(h a, k) and (h + a, k)	(0, -a) and (0, a)	(h, k a) and (h, k + a)
Coordinates of co-vertices	(0, -b) and (0, b)	(h, k b) and (h, k + b)	(-b, 0) and (b, 0)	(h b, k) and (h + b, k)
Coordinates of foci	(-c, 0) and (c, 0)	(h c, k) and (h + c, k)	(0, -c) and (0, c)	(h, k c) and (h, k + c)
Value of c	$c^{2}=a^{2}-b^{2}$	$c^{2}=a^{2}-b^{2}$	$c^{2}=a^{2}-b^{2}$	$c^{2}=a^{2}-b^{2}$

Although the table seems to have a lot of information, there are many parallels between the columns. The vertices (endpoints) are along the major axis or longer axis; the co-vertices (also endpoints) are along the minor axis or shorter axis. Both the major and minor axes are axes of symmetry for the ellipse; the center of the ellipse is at the intersection of the major axis and minor axis. The foci are always along the major axis and are the two points that play a part in the definition of an ellipse. The larger denominator is paired with the variable that indicates the major axis. Thus, if the larger denominator is connected with the variable $x$ , then the major axis is along the $x$ -axis or parallel to the $x$ -axis so that the ellipse is longer horizontally. If the larger denominator is connected with the variable $y$ , then the major axis is along the $y$ -axis or parallel to the $y$ -axis, and the ellipse is longer vertically. If $a=b$ , then the ellipse becomes a circle.

Given an ellipse with center at $(4,-3)$ , vertices at $(-9,-3)$ and $(17,-3)$ , and foci at $(-1,-3)$ and $(9,-3)$ . How can the equation be determined? Notice that the $y$ -coordinates of the two vertices are the same. This means that the ellipse is longer horizontally and the major axis is parallel to the $x$ -axis. The length of the major axis is $17-(-9)=26$ , so $a=13$ . The distance from the center to the foci is 5, so $c=5$ ; using $a=13$ and $c=5$ leads to $b=12$ . Because the center is at $(4,-3)$ , this means $h=4$ and $k=-3$ . Using all this information leads to an equation for the ellipse of $\frac{(x-4)^{2}}{169}+\frac{(y+3)^{2}}{144}=1$ .

Similar analyses can be applied to a hyperbola. A hyperbola consists of those ordered pairs $(x, y)$ in the plane such that the difference of their distances from two fixed points, the foci, is a positive constant.

The table below describes the critical components of a hyperbola. A careful look at this table shows many analogies to the comparable table for the ellipse.

As with an ellipse, a hyperbola has two axes of symmetry which are the transverse axis and the conjugate axis. The foci are along the transverse axis and the vertices are at the endpoints of the transverse axis; the vertices of the transverse axis are the vertices of the hyperbola. The conjugate axis is perpendicular to the transverse axis, and the co-vertices are the endpoints of the conjugate axis; the co-vertices are not points on the hyperbola. The transverse and conjugate axes intersect at the center of the hyperbola. A hyperbola also has two asymptotes that intersect at the center of the hyperbola.

HYPERBOLAS	transverse axis on x-axis	transverse axis parallel to x-axis	transverse axis on y-axis	transverse axis parallel to y-axis
Equation	$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$	$\begin{aligned}&\frac{(x-h)^{2}}{a^{2}} \\&-\frac{(y-k)^{2}}{b^{2}}=1\end{aligned}$	$\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$	$\frac{(y-k)^{2}}{a^{2}}$ $-\frac{(x-h)^{2}}{b^{2}}=1$
Center	(0, 0)	(h, k)	(0, 0)	(h, k)
Length of transverse axis	2a	2a	2a	2a
Length of conjugate axis	2b	2b	2b	2b
Coordinates of vertices	(-a, 0) and (a, 0)	(h a, k) and (h + a, k)	(0, -a) and (0, a)	(h, k a) and (h, k + a)
Coordinates of co-vertices	(0, -b) and (0, b)	(h, k b) and (h, k + b)	(-b, 0) and (b, 0)	(h b, k) and (h + b, k)
Coordinates of foci	(-c, 0) and (c, 0)	(h c, k) and (h + c, k)	(0, -c) and (0, c)	(h, k c) and (h, k + c)
Value of c	$c^{2}=a^{2}+b^{2}$	$c^{2}=a^{2}+b^{2}$	$c^{2}=a^{2}+b^{2}$	$c^{2}=a^{2}+b^{2}$
Equation of asymptotes	$y=\pm \frac{b}{a} x$	$\begin{aligned} y=\pm \frac{b}{a}(x-h) & \\+k \end{aligned}$	$y=\pm \frac{b}{a} x$	$\begin{aligned} y=\pm \frac{b}{a}(x-h) & \\+k \end{aligned}$
Opens how	left and right	left and right	up and down	up and down

Given a hyperbola with center at $(0,0)$ , vertices at $(0,-8)$ and $(0,8)$ , and conjugate axis of length 6. To find an equation for this hyperbola, first note that the $x$ -coordinates of the vertices are the same, so the hyperbola is oriented with the transverse axis along the $y$ -axis. This means the hyperbola opens up and down. The coordinates of the vertices yield $a=8$ . The length of the conjugate axis is $2 b=6$ , so $b=3$ . Putting all these pieces together gives an equation of $\frac{y^{2}}{64}-\frac{x^{2}}{9}=1$ .

The final curve to be studied here is the parabola. A parabola is the set of all ordered pairs $(x, y)$ in the plane that are at the same distance from a fixed line, called the directrix, and a fixed point not on the line, called the focus. As noted in objectives $5 a$ and $5 b$ , a parabola has an axis of symmetry that goes through the vertex; recall that the vertex is the minimum or maximum point of the parabola. Although the focus is not a point on the parabola, there is a segment containing the focus that is sometimes of importance. The latus rectum is a segment parallel to the directrix that contains the focus; the endpoints of the latus rectum are on the parabola.

The table below contains an analysis of the critical components of a parabola similar to those for an ellipse and a hyperbola.

PARABOLAS	axis of symmetry y-axis	axis of symmetry x-axis	axis of symmetry x = h	axis of symmetry y = k
Equation	$x^{2}=4 p y$	$y^{2}=4 p x$	$(x-h)^{2}=4 p(y-k)$	$(y-k)^{2}=4 p(x-h)$
Coordinates of vertex	(0, 0)	(0, 0)	(h, k)	(h, k)
Coordinates of focus	(0, p)	(p, 0)	(h, k + p)	(h + p, k)
Equation of axis of symmetry	x = 0	y = 0	x = h	y = k
Equation of directrix	y = -p	x = -p	y = k - p	x = h - p
Coordinates of endpoints of latus recturm	(-2p, p) and (2p, p)	(p, -2p) and (p, 2p)	(h 2p, k + p) and (h + 2p, k + p)	(h + p, k 2p) and (h + p, k + 2p)
Opens how	up if p > 0 down if p < 0	to right if p > 0 to left if p < 0	up if p > 0 down if p < 0	to right if p > 0 to left if p < 0

Given a parabola with vertex $(2, \, -5)$ and directrix $y=-8$ . Because the equation of the directrix is a horizontal line, the parabola opens either up or down. The equation of the directrix leads to $-8=-5-p$ , so $p=3$ ; because $p > 0$ , the parabola opens up. The equation is then $(x-2)^{2}=12(y+5)$ .

Review this material in Writing Equations of Ellipses, Writing Equations of Hyperbolas, and Parabolas Centered at the Origin.

10b. graph ellipses, hyperbolas, and parabolas centered on and off the origin

Compare the graphs of ellipses, hyperbolas, and parabolas centered at the origin.
How do the graphs of ellipses, hyperbolas, and parabolas transform when the center is not at the origin?

To graph an equation whose graph is an ellipse, hyperbola, or parabola, it is important to start by analyzing the form of the equation. If both $x$ and $y$ are to the second power and are added, the graph is an ellipse; if both $x$ and $y$ are to the second power and are subtracted, the graph is a hyperbola; if only one of $x$ or $y$ is to the second power, the graph is a parabola. If the graph is an ellipse, determine whether the major axis is horizontal or vertical; then identify coordinates of any vertices or co-vertices, and the coordinates of the center. If the graph is a hyperbola, determine whether the transverse axis is horizontal or vertical to know whether the hyperbola opens left/right or up/down, respectively; use the coordinates of the vertices and the center and the equation of the asymptotes to complete the graph. If the graph is a parabola, determine the axis of symmetry to decide if the parabola opens up/down or left/right; then find the coordinates of the vertex and possibly the endpoints of the latus rectum to complete the graph.

Given the equation $\frac{x^{2}}{25}+\frac{y^{2}}{36}=1$ to be graphed. Both $x$ and $y$ are to the second power and are added, so the equation represents an ellipse; the larger number is connected to the $y$ variable, so the ellipse is longer in the vertical direction. No number is subtracted from either $x$ or $y$ , so the ellipse is centered at the origin $(0, \,0)$ . Because $36=6^{2}$ and $25=5^{2}, a=6$ and $b=5$ . Then, the vertices have coordinates $(0, \,-6)$ and $(0, \, 6)$ and the co-vertices have coordinates $(-5, \,0)$ and $(5, \,0)$ . The graph is shown below.

To graph the equation $(x-1)^{2}-\frac{(y+3)^{2}}{16}=1$ , notice that both $x$ and $y$ are to the second power and the variables are subtracted. So, the graph is that of a hyperbola with center at $(1,-3)$ ; the order of the variables indicates that the transverse axis is parallel to the $x$ -axis. The denominator connected to the $x$ -variable is 1, so $a=1$ ; likewise, $b^{2}=16$ , so $b=4$ . So, the coordinates of the vertices are $(0,-3)$ and $(2, \, -3)$ . The graph is shown below. The asymptotes in red have equations $y=\pm 4(x-1)-3$ ; the two asymptotes intersect at $(1$ , -3) which is the center of the hyperbola.

Given the equation $y^{2}=8 x$ to be graphed. Notice that only one of the variables is to the second power, so the equation is of the form for a parabola. Because $y$ is the squared variable, the parabola opens to the left or the right. The vertex is at $(0, \, 0)$ because no transformations are applied to either variable; the axis of symmetry has equation $y=0$ . The coefficient of 8 leads to $8=4 p$ , or $p=2$ ; because $p > 0$ , the parabola opens to the right. The latus rectum has endpoints $(2, \, -4)$ and $(2, \, 4)$ . The graph of the parabola is shown below. The directrix in red has equation $x=-2$ .