9.6 Translate and Classify Conic Sections

 

Conic sections

The intersection of a plane and a double-napped cone

 

General second-degree equation

An equation of the form Ax2 + Bxy + Cy2 + Dx + Ey + F = 0

 

Discriminant

The expression B2 - 4AC for the equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0, used to identify a conic section

 

STANDARD FORM OF EQUATIONS OF TRANSLATED CONICS

In the following equations, the point (h, k) is the vertex of the parabola and the center of the other conies.

Circle        (x - h)2 + (y - k)2 = r2

 

Parabola    (y - k)2 = 4p(x - h)             Horizontal axis

(x - h)2 = 4p(y - k)             Vertical axis

 


Ellipse                                                   Horizontal axis

                                                              Vertical axis

 

Hyperbola                                             Horizontal axis

                                                              Vertical axis

 

Example 1

Graph the equation of a translated circle

 

Graph (x + 3)2 + (y - 2)2 = 4.

1.       Compare the given equation to the standard form of an equation of a circle. The graph is a circle with center at (h, k) = (_-3_, _2_) and radius r =         = _2_.

 

2.        Plot the center. Then plot several points that are each _2_ units from the center:

_(-3 + 2, 2)_ = _(-1, 2)_

_(-3 - 2, 2)_ = _(-5, 2)_

_(-3, 2 + 2)_ = _(-3, 4)_

_(-3, 2 - 2)_ = _(-3, 0)_

 

 

 

3.        Draw a circle through the points.

 

 


Example 2

Graph the equation of a translated hyperbola

Graph

1.         Compare the given equation to the standard forms of equations of hyperbolas. The graph is a hyperbola with a _vertical_ transverse axis. The center is at (h, k) = (_1_, _-2_). Because a2 = _16_ and b2 = _4_, you know that a = _4_ and b = _2_.

 


2.          Plot the center, vertices, and foci. The vertices lie a = _4_ units above and below the center, at (_1__, _2_) and (_1__, _-6__). Because c2 = a2 + b2 = __20__, the foci lie c =  » _4.47_ units above and below the center, at (_1_, _2.47_) and
(_1_, _-6.47_).

3.           Draw the hyperbola. Draw a rectangle centered at (_1_, _-2_) that is 2a = _8_ units high and 2b = _4_ units wide. Draw the asymptotes through the opposite corners of the rectangle. Then draw the hyperbola so that it passes through the vertices and approaches the asymptotes.

 

 



 

Example 3

Write an equation of a translated parabola

 

Write an equation of the parabola whose vertex is at (2,1) and whose focus is at (5,1).

1.       Determine the form of the equation. Sketch the parabola. The parabola opens to the _right_ and has the form (y - k)2 = 4p(x - h) where p _>_ 0.

 

2.        Identify h and k. The vertex is at (2,1), so h = _2_ and k = _1_.

 

3.        Find p. The vertex (2, 1) and focus (5,1) both lie on the line _y = 1_, so the distance between them is |p| = |_5 - 2_| = _3_, and p = _3_ or p = _-3_. Because p _>_ 0, it follows that p = _3_, so 4p = _12_.

 


The equation is _(y - 1)2 = 12(x - 2)_.

 

 

Example 4

Write an equation of a translated ellipse

 

Write an equation of the ellipse with foci at (-2, 3) and (4, 3) and co-vertices at (1, 4) and (1, 2).

1.       Determine the form of the equation. First sketch the ellipse. The foci lie on the major axis, so the axis is _horizontal_. The equation has the form:

 


2.       Identify h and k by finding the center, which is halfway between the foci (or the co-vertices).

(h, k) =                                  = (_1_, _3_)

 

3.       Find b, the distance between a co-vertex and the center, and c, the distance between a focus and the center:

b = |_4 - 3_| = _1_, c = |_-2 - 1_| = 3.

 

4.        Find a. For an ellipse, a2 = b2 + c2 = _12_ + _32 = _10_, so a =                  .

 


The standard form of the equation is                                            .

 

 

CLASSIFYING CONICS USING THEIR EQUATIONS

Any conic can be described by a general second-degree equation in x and y:

Ax2 + Bxy + Cy2 + Dx + Ey + F = 0.

The expression B2 - 4AC is the discriminant of the conic equation and can be used to identify it.

Discriminant                                                                    Type of Conic

B2 - 4AC _<_ 0, B = 0, and A = C                                          Circle

B2 - 4AC _<_ 0 and either B ¹ 0 or A ¹ C                              Ellipse

B2 - 4AC _=_ 0                                                                       Parabola

B2 - 4AC _>_ 0                                                                       Hyperbola

If B _=_ 0, each axis of the conic is horizontal or vertical.

 

Example 5

Classify a conic

 

Classify the conic given by 2x2 + 2y2 - 5x + 3y - 1 = 0.

 

Solution

Note that A = _2_, B = _0_, and C = _2_, so the value of the discriminant is:
B2 - 4AC = 02 - 4(2)(2) = _-16_.

Because B2 - 4AC _<_ 0 and A _=_ C, the conic is a _circle_.