CPhill

e  =  c  /  a

c  =  √[ a^2  - b^2 ]  =   √[80 - 16]  = √64  =  8

a  =  √80  =  4√5

So

e   =    8   /  [ 4√5]   =   2 / √5    ≈  .89

If  the eccentricity  is   0.....we have a circle...so.....the closer  that the eccentricity  is to 1,  the  more elongated it is 

So....this ellipse  is  more elongated than circular

See here to confirm  this, Jenny  :  https://www.desmos.com/calculator/sij3lqn4ts

We need to first find the eqution for the plane

Let vector AB  =  (1, 0 , -1)

Let vector AC  =   ( 1, - 1, 2 )

We can find the normal vector to the plane by taking the cross-product  of  AB x AC

n  =      i    j     k   i     j

           1   0   -1   1    0   =

           1   -1   2   1   -1

[ i * 0 * - 1  +   j * -1 * 1  + k * 1 * - 1 ]  -  [ k * 0 * 1  + i * -1 * - 1 + j * 2 * 1  ]   =

[  0   -  1j - 1k ] -  [ 0  + 1i + 2j ]   =    - 1i  -3j - 1k

The equation  of the plane  using point  A  is  

-1( x - 0)   -3(y -1)  - 1( z  - 1)  = 0

-1x  - 3y  + 3  - 1z + 1  = 0

-1x - 3y - 1z  + 4  = 0

1x + 3y +1z  - 4  = 0

The distance  between  Q  and the plane is given by 

l   1(-2) + 3(3)  + 1( 4)  - 4   l            l  7  l               7

_______________________  =    ______   =      ___  

  √ [ 1^2  + 3^2 + 1^2  ]                    √11               √11

So  d  =  7

THX, Mathleteig  !!!

We wouldn't necessarily have to use Heron's Formula......

Note that  this is an isosceles triangle  with a base of 6  and sides of  5

The altitude  = √ [side ^2  -  (base /2)^2  ]  = √ [ 5^2 - 3^2 ]

So....the area is......

(1/2) base * height

(1/2) 6 *  √[5^2  - 3^2 ]  =

3 * √16  =

3 * 4  =

12 units^2

We have that

( 3a - 1) + (a^2  + 1)  + (a^2 + 2)   = 16         simplify

2a^2  + 3a  + 2  = 16

2a^2  + 3a  -  14  = 0        factor

(2a + 7) ( a - 2)  =  0

Only the second  linear factor produces a positive for a....

a - 2  = 0

a  = 2

So....the sides of the triangle  are

3(2) - 1  =  5

2^2 + 1  =  5

2^2 + 2  =  6

The semi-perimeter, s,  =  [ 5 + 5 + 6 ] /2   =  8

We can find the area by the use of something known as Heron's Formula :

√ [ s ( s - 5) (s - 5) (s - 6)  ]  =

√ [ 8 * 3 * 3 * 2 ]  =

√ [16 * 9  ]  =

√144 =

12 units^2

Let the vector  v  be   ( 2 - 1, 3 - 1,4 - 2)  =  ( 1 , 2, 2 )

So.....the vector  form of the line  is

(1 + t, 1 +2t, 2 + 2t) =    ( t + 1, 2t + 1, 2t + 2)

So   c  = 2     e  = 2     a   =  1

So    ( c / a , e / a )  =    (   2 / 1 ,   2 / 1)   =    ( 2 , 2) 

Note that  as  x increases by 2, f(x)  increases  by 16

So.....the  absolute  difference  between any two outputs for inputs that are two units apart  is   16

Here's another way

Call  x  the number  of draws  in which 5 marbles  are added

Call  y   the number  of draws  in which  3  marbles  are added

So

5x +  3y  =  21

x     y

0    7

1    not an tnteger

2    not an integer

3    2

4    not an integer

5    y  is negative

We can stop  here

x + y   is minimized   when x  =  3  and y  = 2

So...as EP found.....5 draws  !!!

Nice work, CU!!!

*****

Notice that as  x increases by 1,  f(x)  increases by 8

So.....the difference  (in absolute value terms)   =  8