Question

Find the intersection points, identify the type of equation, and the possible number of solutions.
x^2+y^2-16y+39=0
y^2-x^2-9=0
I'm stuck... ;-;

Answer 1

yuck
i guess we can add them maybe to start?

Answer 2

xD Haha, sure, why not. c:

Answer 3

i get
\[2y^2-16y+25=0\]

Answer 4

For... The first one, right?

Answer 5

i just added in a column

Answer 6

Or combined?

Answer 7

the \(x^2\) terms add to zero, leaving what i wrote

Answer 8

ok that is all wrong, lets try something else

Answer 9

Would solving by substitution or addition/elimination work? .-.

Answer 10

i was trying to use elimination but i got the wrong answers for \(y\) for some reason
now sure why it did not work, maybe i added incorrectly

Answer 11

oh because i am an idiot!

Answer 12

xD Lies.

Answer 13

i wrote
\[2y^2-16y+25=0\] when it should have been
\[2y^2-16y+30=0\]

Answer 14

first of before we go any further, do you see how i got that equation ?

Answer 15

Yhu, added down the column again?

Answer 16

right only this time i did it correctly
now we have an easy quadratic equation to solve

\[2y^2-16y+30=0\] or even easier
\[y^2-8y=15=0\] care to solve it? it factors

Answer 17

Should that be +15?

Answer 18

yeah i must be tired
\[y^2-8y+15=0\]

Answer 19

Haha
would it be \[(y+3)^{2}(y-5y)^{2}\]?

Answer 20

?? i thought i was tired
no it would be \((y-3)(y-5)=0\)

Answer 21

; - ;
Can yhu explain why?

Answer 22

you mean can i explain how to factor?
no, you just have to grind it til you find it
but when you have
\[y^2-8y+15=0\] you are looking for two numbers whose product is \(15\) and whose sum is \(-8\) and \(-3,-5\) fit that bill nicely

Answer 23

-.- I switched the signs. :c
Thank yhu...
So after we factor it, would yhu solve for y?

Answer 24

yeah and get \(y=3,y=5\) pretty much in your head
now we have to go back and find the x values for both
there will be four solutions all together, two for \(y=3\) and two for \(y=5\)

Answer 25

OS froze. ;c

So we would have to solve it 4 times?

Answer 26

lets use this one \[y^2-x^2-9=\] with \(y =3\) you get
\[9-x^2-9=0\\
x^2=0\\
x=0\] so \((0,3)\) is one of them

Answer 27

no you don't have to use both equations, only one

Answer 28

but you have to solve it when \(y=3\) and \(y=5\)
\(y=3\) only gives on solution, not two, because \(x^2=0\) only has one solution
now lets try with \(y=5\)

Answer 29

\[y^2-x^2-9=0\\
25-x^2-9=0\\
x^2=16\\
x=\pm4\]

Answer 30

that wasn't too bad

Answer 31

And then we'd plug in the values we just got for x to solve for y? c:

Answer 32

lol no dear we have already solved for y
it is \(y=3\) or \(y=5\)

Answer 33

Maybe I'm tired too. >~<

Answer 34

we did that first
then if \(y=3\) we found that \(x=0\) so one solution is \((0,3)\)

Answer 35

So we come out with 3 solutions...

Answer 36

yeah me too
but to finish it we found it \(y=5\) then \(x=4\) or \(x=-4\) making two more solutions:\[(4,5),(-4,5)\]

Answer 37

Total.

Answer 38

yup 3 total

Answer 39

And one last part for the question, Which type of equations are these?
If I say my guess yhu can't laugh, I don't know if it's right. ; 3;

Would it be a system of equations?

Answer 40

lololololoooollll etc etc

Answer 41

no

Answer 42

one is a circle, the other a hyperbola

Answer 43

\[x^2+y^2-16y+39=0\] is a circle \[y^2-x^2-9=0\] is a hyperbola

Answer 44

I have one I did earlier... Would yhu mind telling me which is which? I got the answersm it was a little easier than this, but I don't know what they are...

\[\frac{ x^{2} }{ 4 }+y^{2}=1\]\[y=x+1\]
I was thinking maybe hyperbola... and line?

Answer 45

a line for sure

Answer 46

the first is not a hyperbola as it is
\[\frac{x^2}{4}\color{red}+y^2=1\]

Answer 47

it is an ellipse
if it had a minus sign there it would have been a hyperbola

Answer 48

Oh... I get those two mixed a lot. Haha, well thank yhu for everything! n.n
Yhur awesome.
And yhu should probably get some sleep. Haha

Answer 49

yw
gnight
you too

Answer 50

'Night c: