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# Hyperbolas Help (page 3)

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By McGraw-Hill Professional
Updated on Oct 4, 2011

## Hyperbola Practice Problems

#### Practice

1. Find the center, vertices, foci, and asymptotes for

2. Find the center, vertices, foci, and asymptotes for
3. Find an equation for the hyperbola having vertices (−4, 2) and (12, 2) and foci (−6, 2) and (14, 2).
4. Find an equation for the hyperbola having vertices (−8, 0) and (−4, 0) and with an asymptote .

In Problems 5–7, match the graphs in Figures 12.28–12.30 with their equations.

1. ( y − 1) 2 − ( x − 1) 2 = 1

2. ( x − 1) 2 − ( y − 1) 2 = 1

Fig. 12.28

Fig. 12.29

Fig. 12.30

#### Solutions

1. h = 5, k = 0, a = 4, b = 3, and

Center: (5, 0)

Vertices: (h, ka ) = (5, 0 − 4) = (5, −4) and (h, k + a) = (5, 0 + 4) = (5, 4)

Foci: ( h , kc ) = (5, 0−5) = (5, −5) and ( h , k + c ) = (5, 0 + 5) = (5, 5)

Asymptotes: and

2. h = −8, k = −6, a = 7, b = 24, and

Center: (−8, −6)

Vertices: ( ha , k ) = (−8 − 7, −6) = (−15, −6) and ( h + a , k ) = (−8 + 7, −6) = (−1, −6)

Foci: ( hc , k ) = (−8 − 25, −6) = (−33, −6) and ( h + c , k ) = (−8 + 25, −6) = (17, −6)

Asymptotes: and

3. The vertices are (−4, 2) and (12, 2), which gives us k = 2 and ( ha , k ) = (−4, 2) and ( h + a , k ) = (12, 2).

A focus is (−6, 2), which gives us ( hc , k ) = (−6, 2) and hc = 4 − c = −6. Solving 4 − c = −6 gives us c = 10. We can find b by letting a = 8 and c = 10 in .

The equation is

4. ( ha , k ) = (−8, 0) and ( h + a , k ) = (−4, 0), so k = 0 and we have the following system.

The slope of an asymptote is , so . The equation is

5. Figure 12.30

6. Figure 12.28

7. Figure 12.29

## Conics on a Graphing Calculator

In order to use a graphing calculator to graph a conic section, the equation probably needs to be entered as two separate functions. For example, the graph of y 2 = x could be entered as . To use a graphing calculator to graph a conic section that is not a function, solve for y . When taking the square root of both sides, we use a “±” symbol on one of the sides. It is this sign that gives us two separate equations.

Solve for y .

•

## Conic Equations - General and Standard Forms

Equations of conic sections do not always come in the convenient forms we have been using. Sometimes they come in the general form Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0. When A and C are equal (and B = 0), the graph is a circle. If A and C are positive and not equal (and B = 0), the graph is an ellipse. If A and C have different signs (and B = 0), the graph is a hyperbola. If only one of A or C is nonzero (and B = 0), the graph is a parabola. There are some conic sections whose entire graph is one point. These are called degenerate conics . We can rewrite an equation in the general form in the standard form (the form we have been using) by completing the square.

#### Examples

Rewrite the equation in standard form.

• x 2 − 2 x − 4 y = 11
• Because there is no y 2 term, the graph of this equation is a parabola that opens up or down. The standard equation is ( xh ) 2 = 4 p ( yk ). We need to have the x terms on one side of the equation and the other terms on the other side.

• −9 x 2 + 16 y 2 − 18 x − 160 y + 247 = 0
• Because the signs on x 2 and y 2 are different, the graph of this equation is a hyperbola. The standard form for this equation is .

## Conic Equations Practice Problems

#### Practice

1. Solve for y

2. Solve for y

3. Rewrite the equation in standard form: 36 x 2 + 9 y 2 − 216 x − 72 y + 144 = 0.

#### Solutions

Practice problems for this concept can be found at: Conic Sections Practice Test.

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