Section 10-6
Rectangular and Parametric Forms of
Conic Sections
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
The graph of a second-degree equation in two
variables always represents a conic or
degenerate case, unless the equation has no
graph at all in the real number plane.
To identify a conic in general form, look at the
coefficients





Identify the conic section represented by each
equation.
1. 3x 2-2y2 -4x +5y = -3 A and C have opposite signs
2
2
2. -4x -4y +6x +y=-5 A = C
circle
2
2
3. 8x + y +3x -7y = 0 A ≠ C b u t th ey h av e th e sam e sig n
2
4. y –x +3y -9 = 0 A= 0
parabola
hyperbola
ellip se



So far we have discussed equations in their rectangular form.
Some conic sections can also be described parametrically.
The general form for a set of parametric equations is x= F(t) and
y= g(t) where t is some interval I. As t varies over I in some
order, a curve containing points (x,y) is traced out in a certain
direction.
A parametric equation can be transformed into its more familiar
rectangular form by eliminating the parameter t from the
parametric equations.
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

Graph the curve defined by the parametric equations x = t+ 1
and y = ± 4t 2 , where -2 ≤ t ≤ 2. T h en id en tify th e cu rve b y
finding the corresponding rectangular equation.
Make a table of values and then graph the curve:
t
x
y
(x,y)
2
(-1, 0 )
-2
-1
0
-1
0
±
0
1
±2
(1, ±2)
1
2
± 3
(2, ± 3 )
2
3
0
3
(0, ± 3 )
-1
3
(3,0)
It appears to be a circle. To identify it accurately, we need to find
the rectangular equation.

Solve the equation x = t +1 for t.
t = x-1. Now substitute it in to the second
equation and square both sides.
2
2
2
2
y = 4 - (x-1) = 4 – x + 2x -1 = -x +2x +3
2
2
x + y - 2x = 3
2
2
(x -1) + (y-0) = 4 center (1,0) and radius 2
Some parametric equations require the use of
trigonometric identities to eliminate the
parameter t.

Find the rectangular equation of the curve whose parametric
equations are x = - cos t and y= sin 2 t , w h ere 0 d eg rees ≤ t ≤ 180
degrees. Then graph the equation using arrows to indicate how
the graph is traced.
cos t = -x sin t = ± y
Now use the trigonometric identity cos2 t + sin2 t = 1 to rewrite the
equation to eliminate t.
2
2
2
(-x) + y = 1
x = -y + 1 (parabola) ( x + 0) = 4(-.25)(y-1)

Vertex (0, 1)




t
x
y
(x,y)
0 degrees
-1
0
(-1,0)
60
-.5
.75
90
0
1
(0,1)
180
1
0
(1,0)



1
^
(-.5,.75)
-1
1



Find the parametric equations for the equation
2
2
x/9 + y/4 = 1
2
2
We know cos t + sin t = 1 substitute in
2
2
2
2
x/9 = cos t and y/4 = sin t so x = 3 cos t and
y = 2 sin t 0≤ t ≤ 2Л
 Section
10-6
 Pp. 667-669
 #13-23 odds, 45,46,47