## Sunday, January 22, 2017

### Solutions to Analytic Geometry Problems (2)

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 +    y2  / 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 x2  +   9  y2   = 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 x2  +    y2   =    1

Solution for this graph:

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

Then: c  =   Ö (a 2  -   b 2)   =      Ö  (1   -  1/4)   =   Ö  (3/4)   =  Ö 3/  2

eccentricity e =   c/  a   =  Ö 3/  2   /  1  =  0.866

5) Look up the definition of the "directrix" of an ellipse. Then find the eccentricity and directrices of the ellipse:

x2 /7 +    y2   / 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 semi-major axis.

The graph is shown below:

Let   a  =  Ö 7

i.e. given the general form for ellipse equation: x2 / a 2 +    y2  / b 2 = 1

then  b =  4   (this is now the semi-major axis since b > a)

=   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.