CPhill

Simplify as

2x^2 + 4x  + 4y^2  + 20y   =  -k    complete the square on  x and  y

2(x^2 + 2x  + 1)  + 4(y^2 + 5y + 25/4)  =  -k  + 2  + 25

2 ( x + 1)^2   +  4( y + 5/2)^2  =  27  - k

(x + 1)^2 / (1/2)  + (y + 5/2)^2  / (1/4)  = 27  - k     divide both sides by (27 -k)

( x + 1)^2 / [ (1/2)(27 -k)]  +  (y+ 5/2)^2 / [(1/4) (27 -k) ]  =  1

a^2  =  (27 - k) / 2       b^2  = (27 - k) / 4

a =  sqrt [(27 -k) /2 ]    b = sqrt [ (27 -k) /4)]

Area =  pi * a * b   =  6  pi ......so

pi * a * b   = 6 pi

  a *b   = 6

sqrt  [ (27 -k) / 2   *  (27 -k) /4 ]  =  6

(27 - k) / sqrt 8  =  6

(27 - k) / (2sqrt 2) = 6

27 -k  =  6 * 2sqrt 2

27 - k  = 12sqrt 2

k  =  27 - 12sqrt 2

Using polynomial division

                          x^2 - 2x + 8

x^2  + 2x  - 5  [  x^4  + 0x^3  - x^2  + 3x  -7 ]

                          x^4   +2x^3  - 5x^2

                         _______________________

                                   -2x^3  + 4x^2  + 3x

                                  -2x^3   - 4x^2  + 10x

                                 ___________________

                                               8x^2   - 7x  - 7

                                               8x^2   +16x -40

                                               _____________

                                                        -23x  + 33

Q = x^2 -2x + 8

R = 33 - 23x

I'll try  !!!

Let A  =(-3,3)  B =  (3,3)   C  =(3,-3)  D = (-3,-3)

Let  the  center of  the ellipse = (0,0) ....let  one of the axis lie along x....such that (a,0)    = P 

The equation of the ellipse is x^2/a^2 + y^2/b^2  =1

Constructing circles at  (3,3)  and (3,-3)  both with radii = 8   we have the equations

(x - 3)^2  + ( y -3)*2   =8^2

(x -3)^2 + (y +3)^2  = 8^2

Let y = 0  and we can  find a  as

(x-3)^2 + 9 = 8^2

(x -3)^2 = 55

x- 3  = sqrt (55)

x = sqrt(55) + 3  = a  ≈  10.42

To find b^2  we can use this equation

x^2 / 10.42^2  +  y^2 / b^2   =1

Since (x,y)  =  (3,3) is on the ellipse  then

3^2 /10.42^2   +  3^2 / b^2  = 1

9 / 10.42^2  + 9/b^2  =1

9 / b^2  =  1 - 9/10.42^2

9/b^2 = .917

b^2  = 9/9.17

b = sqrt (9/.917)  ≈  3.13

The  approximate  area of  the ellipse =  pi * a * b   = pi * 10.42 * 3.13 = 102.46    

\( PA \cdot PB \)

Let the center of  both be (0,0)

Let the ellipse have its major axis along x....its equation is

x^2 / 12^2   + y^2 / (6.5)^2   = 1

The  foci  of the ellipse  =  (+/-  sqrt ( 12^2 - 6.5^5] , 0)  = ( +/- sqrt (407) / 2) , 0)

For  the hyperbola  the equation is

x^2/a^2  - y^2/b^2  =1

a^2 = 2.5^2  

c =  sqrt (407)/2

b^2  =  c^2  - a^2 =  407/4  -  2.5^2  =  191/2

The equation of the hyperbola is

x^2/ 2.5^2  - y^2/ (191/2)  =1

This is  a little difficult to solve  but here's a graphical solution using GeoGebra

PA = 9.5

PB =14.5

PA * PB  =  9.5 * 14.5 =  137.75

The function  can be written as

y = Asin (Bx)  +  C

A= the amplitude

B = 2pi / period =  2pi /pi  =  2

C = the vertical shift  =   -2

So

The function is

y = 3sin (2x) - 2

f(x)  = -3x^2  -6x + 27

Set to 0  to find the roots

-3x^2  -6x + 27 = 0

3x^2 + 6x  -27 = 0

3 (x^2 + 2x - 9) = 0

x^2 + 2x - 9 = 0

x^2  + 2x  =  9    complete the square on x

x^2  + 2x + 1 =  9 + 1

(x + 1)^2    = 10

x + 1  =  sqrt 10          or       x +  1  =  -sqrt 10

x = sqrt 10  -1           or x =  -sqrt 10  - 1

Negative on

(inf , -sqrt (10 )  - 1 )   U  (sqrt (10) -1, inf )

a(1)^2 + b(1) + c =  -24

a(4)^2 + b(4) + c = 10

a(3)^2 + b(3) + c = 60         simplify these

a +  b +  c =  -24           (1)

16a + 4b + c = 10          (2)

9a + 3b + c = 60            (3)

Subtract (2) - (1)  and (3) -(1)

15a + 3b  = 34     (4)

8a + 2b  =  84  →  4a + b  = 42  →  b = 42 - 4a       (5)

Sub (5)  into (4)

15a + 3(42 - 4a)  =  34

15a + 126 -12a = 34

3a  = -92

a = -92/3

b = 42 - 4(-92/3)  = 494/3

a + b + c =  -24

-92/3 + 494/3 + c = -24

c = -24 + 92/3  - 494/3   = -158

f(-1)  = (-92/3)(-1)^2  + (494/3)(-1)  - 158  =   -1060 / 3 

https://web2.0calc.com/questions/sum_93

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