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avatar+1897 

im unsure how to go forth in answering this question, can someone help me?

 Nov 6, 2019
 #1
avatar+1897 
+2

actually need help with this oneas well, and thats it. thank you!

 Nov 6, 2019
 #2
avatar+111357 
+2

Second one, first   [easier to see  ]

 

 

x^2 + y^2  = 9        (1)

 

x^2         y^2

___  -  _____  =      1           (2)

 9            4

 

Multiply  (2)  through  by     4      and we get that

 

(4/9) x^2  -  y^2   =  4         (3)

 

Using (1)  and (3)  we have that

 

x^2         +     y^2    = 9

(4/9)x^2  -      y^2    =  4           add these equations

 

(13/9)x^2  =  13             multiply both sides by   (9/13)

 

x^2  =  9            take both roots

 

x  = -3      or  x   = 3

 

Using  (1)   ....no matter if x  = -3  or  x  = 3.....we have that

 

(3)^2  + y^2   =   9

 

9 + y^2   =  9             subtract   9 from both sides

 

y^2   = 0

 

y  = 0

 

So....the solutions are     (3, 0)   and  (-3,0)

 

The two graphs are  a circle and a hyperbola : https://www.desmos.com/calculator/wwigbxzcsz

 

 

cool cool cool  

 Nov 6, 2019
 #8
avatar+1897 
+1

I will make note of this ! Thanks for explaining it piece by piece, it helps me understand!

jjennylove  Nov 6, 2019
 #3
avatar+111357 
+2

y^2           x^2

___   -    _____    =     1            (1)

 4               2

 

y^2  = 8(x + 1)           (2)

 

Sub (2) into (1)   and we get that

 

8(x + 1)          x^2

_______  -   _____    =   1               Multiply this through by   4 

       4              2

 

 

8(x + 1)  -  2x^2   =    4                 

 

8x +  8   -  2x^2  =  4        rearrange as

 

2x^2  -  8x   -  4  =  0        divide through by 2

 

x^2  -   4x  -  2   = 0          

 

Using the quad formula

 

x =     4 ±√[ 4^2  - 4 (1) (-2) ]                  4  ±√24            4 ±√[4*6]          4 ±2√6

        ______________________  =     ________  =    _________  =    ______   =  2±√6

                       2 * 1                                      2                     2                      2

 

When   x  =    2 + √6   then  we have that

 

y^2  =  8( 2 + √6  + 1)

y^2  =  8 ( 3 + √6)           take both roots

y  =  ±√[ 8 (3 + √6) ]

y = ±2√[ 2 (3 + √6 ]

y  = ±2√[6 + 2√6 ]

 

And when  x  =  2  -√6     we have that

y^2 = 8( 2 - √6  + 1)         take both roots

y  = ±√[ 8 ( 3 - √6) ]

y = ±2√[2(3 - √6) ]

y = ±2√[6 - 2√6 ]

 

So......

When  x  = 2 +√6....then  y = 2√[6 + 2√6 ]    and  -2√[6 + 2√6 ]

When  x = 2 - √6....then y  =   2√[6 - 2√6 ]    and  -2√[6 - 2√6 ]       

 

 

 

cool cool cool

 Nov 6, 2019
 #4
avatar+1897 
+1

hmm, I kind of confused still. I tried looking at your work and then comparing it to the answer choices, but im unsure which ones it would be? these two?

 

jjennylove  Nov 6, 2019
 #5
avatar+111357 
+2

The correct choices are

 

(2 + √6 , 2√[ 6 + 2√6 ]  )

 

(2 + √6 , -2√[ 6 + 2√6 ]  )

 

(2 - √6 , 2√[ 6 - 2√6 ]  )

 

(2 - √6 , -2√[ 6 - 2√6 ]  )

 

 

cool cool cool

 Nov 6, 2019
 #7
avatar+1897 
+1

ohhh okay, I see the answers now . It all makes sense ! Thank you for helping me with these 2 questions!

jjennylove  Nov 6, 2019
 #9
avatar+111357 
+1

OK......hope it made sense  ....!!!!

 

 

cool cool cool

CPhill  Nov 6, 2019
 #6
avatar+111357 
+2

We have four points of intersection....so.....there will be four solutions

 

See the graph here  :  https://www.desmos.com/calculator/lwcv8tjcrq

 

 

cool cool cool

 Nov 6, 2019

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