*Solution*:

We have:

x

^{ }

^{2 }

**/**a

^{2 }+ y

^{ }

^{2 }

^{ }**/**b

^{2}= 1

_{}

^{}

Factor to get:

(x/a + y/b) (x/a - y/b) = 1

Take limit for one of the factors, viz:

lim x

_{ ®oo}(x/a - y/b) = 0

y

_{ }

_{®oo}
Which leads one to conjecture that the straight line:

(x/a - y/ b) = 0 may be an asymptote for the curve.

Rewrite: x/ a = y/ b or y = bx/ a

And we see it comports with the example in the text, e.g.

y = bx/a, or (for specific case) y = -2x/ 3 and y = 2x/ 3.

b) Label (on a graph) and identify the coordinates of the foci for the hyperbola:

y

^{2 }

**/**9

^{2 }- x

^{2 }

^{ }**/**4

^{2 }= 1

Solution:

In this case the two foci (F1 and F2) lie on the

**y-axis**. Obtaining the coordinates of the foci is straightforward and is just: F2 (0, +c) and F1(0, -c) where c =

**Ö**13 or 3.6. So the labeled graph is shown below:

2) In Problem #5 from the ellipse set, we saw an ellipse with equation:

x

^{ }

^{2 }

**/**7 + y

^{2 }

^{ }

**16 = 1**

^{ }/Change this to the form for a hyperbola, and:

a) sketch the graph

b) show the asymptotes

c) identify the foci coordinates

d) If the eccentricity of a hyperbola is expressed

e =

**Ö**(a

^{2}+ b

^{2}) / a

Then label the directrices of the hyperbola in the graph if they are at distances (a/e) from the intersection of the asymptotes.

*Solution*:

We write the equation for the relevant hyperbola:

x

^{ }

^{2 }

**/**7 - y

^{2 }

^{ }

**16 = 1**

^{ }/The graph with the asymptotes and foci show is sketched below:

The foci coordinates are defined at F1 (+c, 0) and F2 (-c, 0) where in this case, c =

**Ö**23 = 4.8.

The foci coordinates then are at: F1 (

**Ö**23 , 0 ) and F2 (-

**Ö**23 , 0)

_{}

^{}

The eccentricity would be: e =

**Ö**(a

^{2}+ b

^{2}) / a

Or: e =

**Ö**23 /

**Ö7**= 1.81

3) Analyze the equation below using the translation of axes approach:

x

^{ }^{2 }**- 4 y**^{ }^{2 }^{ }^{ }- 2 x + 8 y - 2 = 0
Then:

a) Find the equation of the hyperbola with center at x' = y' = 0

b) The intercepts on the y' axis

c) the coordinates of the foci F1 and F2

d) The equations for the directrices

We rewrite the equation:

(x - 1)

Then:

4 (y - 1)

At first one might be tempted to divide through by 4, but this isn't useful because we don't want a fraction on the right side (e.g. 1/4). So better to rewrite the first term as:

4 (y - 1)

Then we have:

(y - 1)

Or:

(y - 1)

By way of translation: x' = x - 1 and y' = y - 1

The translated hyperbola equation becomes:

y'

Which denotes a hyperbola with center at x' = y' = 0 or x = y = 1.

The graph is shown below with asymptotes and foci marked:

The foci F1, F2 are at (0,

where: a = 0.5 and e =

*Solution*:We rewrite the equation:

(x - 1)

^{2 }**- 4 (y - 1)**^{2 }^{ }^{ }= 2 + 1 - 4 = - 1Then:

4 (y - 1)

^{2 }^{ }^{ }- (x - 1)^{2 }**= 1**At first one might be tempted to divide through by 4, but this isn't useful because we don't want a fraction on the right side (e.g. 1/4). So better to rewrite the first term as:

4 (y - 1)

^{2 }= (y - 1)^{2 }/ (1/4) = (y - 1)^{2 }/ 0.25Then we have:

(y - 1)

^{2 }^{ }^{ }/ 0.25 - (x - 1)^{2 }**= 1**Or:

(y - 1)

^{2 }^{ }^{ }/ 0.25 - (x - 1)^{2 }/1 = 1By way of translation: x' = x - 1 and y' = y - 1

The translated hyperbola equation becomes:

y'

^{2 }^{ }^{ }/ 0.25 - x'^{2 }/1 = 1Which denotes a hyperbola with center at x' = y' = 0 or x = y = 1.

The graph is shown below with asymptotes and foci marked:

The foci F1, F2 are at (0,

__+__**Ö1.25)****The equations for the directrices are: x = a/ e and x = - a/e**

where: a = 0.5 and e =

**Ö**(a^{2}+ b^{2}) / a = Ö1.25 / 0.5 = 2.23_{}^{}
Then:

x = 0.5/ 2.23 = 0.22 and x = - 0.22

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