Question

find the intersect point of y^2+x^2-16y+39=0 and y^2-x^2-9=0

Answer 1

looks like you have a circle and a hyperbola right?

Answer 2

maybe we can solve be solving both equations for \(x^2\) as a first step, although there might be an easier way

Answer 3

ok well one is x^2=y^2-9 right?

Answer 4

yes

Answer 5

yes, this works out nicely
how about the other one, you got that?

Answer 6

yeah its x^2=-y^2+16y-39

Answer 7

that is what i get to
now we can solve
\[y^2-9=-y^2+16y-39\]
know how to do that?

Answer 8

*too

Answer 9

yeah one sec

Answer 10

k
suggestion: at one step, divide by 2
also it factors

Answer 11

i did-
y^2-9=-y^2+16y-39 +9 to both sides
y^2=-y^2+16y-30 +y^2 to both sides
2y^2=16y-30 divide both sides by 2
y^2=8y-15
and thats as far as i understand what to do unless i did something wrong?

Answer 12

looks good except that you have a quadratic equation (degree 2) so you want to set it equal to zero, i.e. write
\[y^2-8y+15=0\] in order to solve it
you can factor this one, so the zeros are not too hard to find i.e. you do not need the quadratic formula

Answer 13

clear how to factor this one?

Answer 14

im not very good with factoring

Answer 15

think of two numbers whose product is \(15 \) and whose sum is \(-8\)
not too many choices here

Answer 16

if it is still not clear let me know and i will tell you

Answer 17

3 and 5

Answer 18

(y-3)(y-5)right?

Answer 19

well, actually \(-3\) and \(-5\) fit the bill, since they have to add to \(-8\)
but that means it factors as
\[(x-3)(x-5)=0\] so the actual zero are \(3\) and \(5\)

Answer 20

yes, you are right

Answer 21

oops typo there, i means \(y\) not \(x\) sorry

Answer 22

\[(y-3)(y-5)=0\implies y=3,y=5\]
now we know two values for \(y\) but we still need to find \(x\)

Answer 23

i think that i misinformed you in my question, im looking for the intersection points that the equations have together if not then keep going

Answer 24

yeah i know that we are almost done

Answer 25

oh ok cause you said zeros and i need the point that they intersect eachother lol

Answer 26

i meant the zeros of the quadratic equation we had to solve
we have two possible values for \(y\) namely \(3\) and \(5\)
we can now find the corresponding values of \(x\) because we know
\[x^2=y^2-9\]

Answer 27

if \(y=3\) then this equation becomes \(x^2=3^2-9=0\) and so \(x^2=0\) tells us \(x=0\)
one point of intersection is therefore \((0,3)\)

Answer 28

now we have to be careful not to make a mistake with \(y=5\)
get
\[x^2=5^2-9=25-9=16\] i.e. \[x^2=16\]

Answer 29

what is \(x\) ?

Answer 30

x=4

Answer 31

careful

Answer 32

one possible answer is \(4\) and so one point of intersection is \((4,5)\)
but there is another ...

Answer 33

im confused

Answer 34

\(x^2=16\) has two solutions, not just one

Answer 35

you got one right away: \(x^2=16\) one answer is \(x=4\)
you know the other one? (hint: it is negative)

Answer 36

-4

Answer 37

right!
remember i said be careful here
we did not have to worry about it with the first one, because \(x^2=0\) has only one solution

Answer 38

so there are 3 points of intersection:
\[(-4,5), (0,3),(4,5)\]

Answer 39

wanna check the answer?

Answer 40

i can do it, THANK YOU FOR YOUR HELP!!!

Answer 41

yw my pleasure
i didn't mean actually check as in "plug in the numbers" i meant "cheat"

http://www.wolframalpha.com/input/?i=y^2%2Bx^2-16y%2B39%3D0%2C+y^2-x^2-9%3D0

Answer 42

solve the secold equation for x^2