__Problems:__1) In the last example problem worked above, we saw that the ellipse equation of form:

(x + 2)

^{2}

**/**4 + (y - 1 )

^{2}

^{ }

^{}^{ }

**9 = 1**

^{ }/Could be simplified to: x'

^{2}

**/**4 + y '

^{2}

^{ }

^{}^{ }

**9 = 1**

^{ }/Using: x' = (x + 2) and y' = ( y - 1)

Show how the

*offset*affects the location of the simplified ellipse form by graphing

*both ellipses*on the same axes. Thus, satisfy yourself that the graph shown is justified once the x, y- coordinate offsets are taken into account.

Solution:

The two ellipses graphed on the same axes are shown below:

where the "primed" form of the graph is the one that is centered at (0,0).

2) An ellipse has center coordinates at C(-3,0) and focus at F(-3, -2) with a = 4. Find the eccentricity e and sketch the graph.

We have for coordinate-y difference: (0 - (-2)) = 2 so c = 2

The eccentricity e = c / a = 2/ 4 = 0.5

b =

**Ö**(a^{2}- c^{2}) =**Ö**(16 - 4 ) =**Ö**12
The equation is then (offset 3 units in x coordinate):

(x + 3)

**/**a^{2}+ y^{2}^{ }**b**^{ }/^{2}= 1
Or:

(x + 3)

^{2}**/**16 + y^{2}^{ }**12 = 1 (see below)**^{ }/_{}^{}3) An ellipse has center coordinates at C(2, 2) and focus at F(-1, 2) with a =

**Ö**10. Find the eccentricity e and sketch the graph.

e = 3/

**Ö**10

Graph is:

4) Sketch the following ellipses from their equations below, and give the semi-major axes, semi-minor axis and eccentricity e.

a) 4 x

^{2}

**+ 9 y**

^{2}

^{ }

^{}^{ }= 144

Solution for this graph:

From which we see: a = 6 units, and b = 4 units

Then: c =

**Ö**(a

^{2}- b

^{2}) =

**Ö**(36 - 16) =

**Ö**20

eccentricity e = c/ a =

**Ö**20 / 6 = 0.745

b) 4 x

^{2}

**+ y**

^{2}

^{ }

^{}^{ }= 1

Solution for this graph:

From which we see: a = 1 units, and b = 1/2 unit

Then: c =

eccentricity e = c/ a =

5) Look up the definition of the "

x

And sketch the graph showing the locations of the foci.

Solution:

Each ellipse, with two foci must have two associated directrices. One is shown below:

The graph is shown below:

Let a =

i.e. given the general form for ellipse equation: x

then b = 4 (this is now the

c =

Then: c =

**Ö**(a^{2}- b^{2}) =**Ö**(1 - 1/4) =**Ö**(3/4) =**Ö**3/ 2eccentricity e = c/ a =

**Ö**3/ 2 / 1 = 0.8665) Look up the definition of the "

*directrix*" of an ellipse. Then find the eccentricity and directrices of the ellipse:x

^{2}**/7**+ y^{2}^{ }^{}^{ }**16 = 1**^{ }/And sketch the graph showing the locations of the foci.

Solution:

Each ellipse, with two foci must have two associated directrices. One is shown below:

**Each directrix**being perpendicular to the line joining the two foci then the eccentricity can be interpreted as the position of the focus as a fraction of the semimajor axis.The graph is shown below:

Let a =

**Ö**7i.e. given the general form for ellipse equation: x

^{2}**/**a^{2}+ y^{2}^{ }**b**^{ }/^{2}= 1then b = 4 (this is now the

*semi-major axis*since b > a)c =

**Ö**(b^{2}- a^{2}) =**Ö**(16 - 7) =**Ö**(9) = 3
eccentricity e = 3/4 = 0.75

Directrices at: y =

__+__b

^{2}/ c =

__+__16/ 3

_{}

^{}

__Note__: In working these problems it also helps to realize that the ellipse can't have an e -value greater than 1 (which which would make it a hyberbola) and one also can't have imaginary values for the radical defining c.

## No comments:

Post a Comment