Hi,
How do we solve systems of conic equations? For example:
and
How do we find the point(s) of intersection?
Thanks!
Not sure I've ever heard the term "Conic Equation". Anyway...
First, you should come to grips with the fact that there may be no point of intersection.
After that, a little quick homework can narrow things down quickly. The first, centered at (3,-9) and reaching 4 or 8, depending which way you go, might suggest that it does run into the second, whcih is centered at (9,-2) and reaches 3 or 7, depending on direction.
Next, that ellipse doesn't get anywhere near Quadrant II, so we can just ignore that piece of the hyperbola.
Hyperbola's asymptotes are pretty clear at
y = 2x+3 and y = 15-2x
Again, the ellipse doesn't get all that close to y = 2x+3, so I'm guessing that a new problem is in order. Find the points of intersection between the ellipse and y = 15-2x. This WILL tell you where to start looking. One should be closer than the other, it is an asymptote.
Now what?
What TKHunny is saying is that how you determine a solution will depend strongly upon the specific conics. For these specific equations one obvioius method is to solve the first equation for y:
so
Now replace y in the second equation by that. You will eventually get a fourth degree equation.
Corrected thanks to Wilmer.
Did you realise that those are 2 ellipses, IF you change 1st equation to "positive"?
The basic equation of an ellipse is x^2 / a^2 + y^2 / b^2 = 1 ; a > b
If you're "learning", why are you starting with a complicated case?
Did you google"ellipse equation"? Did you graph an ellipse to "see"?
Good "lessons" here:
http://www.solitaryroad.com/c428.html
I sincerely hope I did not say that. One needs a sense of it. Unfortunately, for the discussion above, I seem to have graphed badly and put the ellipse in the wrong place. Such is one of the dangers of graphing. Once you SEE it, even if it is wrong, it is difficult to get out of your head.
Solving such problems is one of the joys of pre-calculus mathematics. I recommend finding ways to solve without the super-duper brute force suggested by HallsofIvy. For the record, this would also be my first attempt, generally, but this one seems like we can do something else with it. There's just too much symmetry to let it go.
I would try this, first, now that I have abandoned my dear old friend, Brute Force.
Solve the ellipse for (x-3)^2 and substitute into the hyperbola. See if you can get it. (x-3)^2 = 12x - (1/49)[3123+36y+9y^2].
This substitution turns the hyperbola into: 2352x = 19597 + 1026y + 85y^2
If you also solve the hyperbola for (y+2)^2, getting (y+2)^2 = 4x^2 - 24x - 14y -105
and make a similar substitution into the ellipse, the ellipse reduces to 126y = 2583 - 1098x + 85x^2
Now, it's almost easy. Who can't find the intersections of two parabolas?
To really get familiar with "intersecting points", I suggest you try 2 ellipses; as example:
x^2 / 36 + y^2 / 9 = 1 (horizontal)
x^2 / 4 + y^2 / 25 = 1 (vertical)
You should end up with 4 intersecting points:
[4/sqrt(6), 5/sqrt(3)], [-4/sqrt(6), 5/sqrt(3)], [-4/sqrt(6), -5/sqrt(3)], [4/sqrt(6), -5/sqrt(3)]
And you can actually "have fun" doing the work!!
The first equation is a hyperbola. I'm going to assume multiplying both sides by -1 is what you mean by "[changing] the 1st equation to 'positive'". Even if you multiply the equation by -1, you still won't get an ellipse, because the 1 on the right side will become negative.