Question

How would I go about finding the intersections of these two equations? (The first one is a circle and the second is a parabola)

(posting equations in comments)

Answer 1

\[x ^{2}+y ^{2}-16y+39=0\]\[y ^{2}-x ^{2}-9=0\]

Answer 2

make x the subject of the 2nd equation and insert into the 1st

Answer 3

so x^2 = y^2 - 9 ?

Answer 4

yh, sub into 1st

Answer 5

Tammi: Please hold it. Check your second equation. I think that's the equation of a hyperbola, not a parabola.

Answer 6

Oh yes, it is! My bad

Answer 7

And how do I sub into 1st...

Answer 8

Tammi: Check that 2nd again, please. Solve for y^2. Doesn't matter all that much which oone you solve for, x^2 or y^2, but please solve for y^2 here.

Answer 9

x^2 +y^2 -16y +39=0 and x^2 = y^2 -9 so (y^2-9) + y^2 - 16y + 39 = 0

Answer 10

so then -y^2 = -x^2 - 9 ? I don't know how to get rid of the exponent

Answer 11

Take a moment and ask y ourself why you're solving the 2nd equation for y^2 and why you're substituting the result into the first equation.

Answer 12

Because there are two ys in that equation and one x ?

Answer 13

The reason is that substitution eliminates one variable.
From the second equation you could get either -x^2=9-y^2 or y^2=9+x^2. Look carefully at the 1st equation presented in this problem and ask yourself which you

Answer 14

should substitute, y^2=9+x^2 or x^2+y^2-9. Sorry this took so long.

Answer 15

I'd substitute for y^2 because x^2 is not negative...?

So the equation would become \[x ^{2}+(9+x ^{2})-16y+39=0\]?

Answer 16

Thanks for using the equation editor. As before, my point is that substitution can enable you to eliminate one variable. In the equation y ou've just typed, you still have both x and y in the equation.

Answer 17

So my suggestion is to take the second equation and transform it into x^2=y^2-9.
If you substitute that into the first equation, x will disappear completely and you'll be left with only y in your equation. How much sense does this make?

Answer 18

That makes perfect sense, so\[(y ^{2}-9)+y ^{2}-16y+39=0\]

Answer 19

That's really great; I see you know your stuff. Now would y ou please combine those two y^2 terms and type out the equation as it now remains?

Answer 20

\[2y ^{2}-16y+30=0\]

Answer 21

Cool, great work.
Divide all 3 terms by 2 to siimplify the equation, then type out the new form of the equation.

Answer 22

\[y ^{2}-8y+15=0\]

Answer 23

Draw on your past experience to plan what you need to do next to solve for y.

Answer 24

factor it?

Answer 25

That might work, and if it does, your suggestion is better than what I had had in mind.
Would you try factoring Y^2-8y+15=0.

Answer 26

(y - 3)(y - 5)
(y * y)(y * -5)(-3 * y)(-3 * -5)\[y ^{2}-5y-3y+15\]\[y ^{2}-8y+15\]So (y - 3)(y - 5) are factors

Answer 27

Great, and so what are the two solutions (y-values)?

Answer 28

y = 3 and y = 5

Answer 29

Yes! Now, if you'd please go back to the 2nd equation presented in this problem, and substitute y=3, you'll find that x=0

Answer 30

What happens if you substitute y=5? You'll get a different answer.

Answer 31

So, Tammi: if y=3, x=0 and y our solution therre is (0,3).
And if y=5, x = ??? and so you'll have 2 more solutions: (???, 5) and (???, 5).
Does this make sense to you? As much as I've enjoyed

Answer 32

working with you, I need to leave. You could check y our results by substituion into the 1st or the origianl equations.

Answer 33

\[(5)^{2}-x ^{2}-9=0\]\[25 - x ^{2}-9=0\]\[x ^{2} + 16 = 0\]\[x ^{2}=-16\]\[x = \pm4\]?

Answer 34

Actually, Tammi, you'd get -x^2+16=0. Thus, x would be plus or minus 4, as you have it.

Answer 35

Oh okay c:

Answer 36

So, are you OK, are y ou satisfied, or what? Strongly suggest your check your answers by subsitution of the 3 points back into the original (top) equation.

Answer 37

Yes, I'm in the process of doing that c: I'll post my final answers when I'm done, but I'm okay now (: thank you

Answer 38

Best of luck to you, Tammi. You know a lot already, and I respect you for it.